# Some Basic Concepts of Chemistry 

### 1.3. Properties of Matter and their Measurement

1. Weight (g) of two moles of the organic compound, which is obtained by heating sodium ethanoate with sodium hydroxide in presence of calcium oxide is:
(A) 30
(B) 18
(C) 16
(D) 32
[NEET 2023]
2. 0.24 g of a volatile gas, upon vapourisation, gives 45 mL vapour at NTP. What will be the vapour density of the substance? (Density of $\mathrm{H}_{2}=0.089$ )
(A) 95.93
(B) 59.93
(C) 95.39
(D) 5.993
[AIPMT 1996]
3. The dimensions of pressure are the same as that of:
(A) force per unit volume
(B) energy per unit volume
(C) force
(D) energy
[AIPMT 1995]

### 1.4. Uncertainty in Measurement

4. Given the numbers: $161 \mathrm{~cm}, 0.161 \mathrm{~cm}, 0.0161 \mathrm{~cm}$. The number of significant figures for the three numbers are:
(A) 3,3 and 4 respectively
(B) 3, 4 and 4 respectively
(C) 3, 4 and 5 respectively
(D) 3, 3 and 3 respectively
[AIPMT 1998]
5. In the final answer of the expression.

$$
\frac{(29.2-20.2)\left(1.79 \times 10^{5}\right)}{1.37}
$$

the number of significant figures is:
(A) 1
(B) 2
(C) 3
(D) 4
[AIPMT 1994]

### 1.5. Laws of Chemical Combinations

6. Equal masses of $\mathrm{H}_{2}, \mathrm{O}_{2}$ and methane have been taken in a container of volume V at temperature $27^{\circ} \mathrm{C}$ in identical
(C) The topic number and names correspond to the exact topics of the NCERT Class XI textbook, so it's easy for you to find in NCERT.
condition. The ratio of the volumes of gases $\mathrm{H}_{2}: \mathrm{O}_{2}: \mathrm{CH}_{4}$ would be:
(A) $8: 16: 1$
(B) $16: 8: 1$
(C) $16: 1: 2$
(D) $8: 1: 2$
[NEET 2016]
7. The molecular weight of $\mathrm{O}_{2}$ and $\mathrm{SO}_{2}$ are 32 g and 64 g respectively. At $15^{\circ} \mathrm{C}$ and 150 mmHg pressure, one litre of $\mathrm{O}_{2}$ contains ' $N$ ' molecules. The number of molecules in two litres of $\mathrm{SO}_{2}$ under the same conditions of temperature and pressure will be:
(A) $N / 2$
(B) $N$
(C) $2 N$
(D) $4 N$
[AIPMT 1990]

### 1.7. Atomic and Molecular Masses

8. Suppose the elements $X$ and $Y$ combine to form two compounds $X Y_{2}$ and $X_{3} Y_{2}$. When 0.1 mole of $X Y_{2}$ weighs 10 g and 0.05 mole of $X_{3} Y_{2}$ weighs 9 g , the atomic weights of $X$ and $Y$ are:
(A) 40,30
(B) 60,40
(C) 20,30
(D) 30,20
[NEET Phase II 2016]
9. An element $X$ has the following isotopic composition:

$$
{ }^{200} X: 90 \% \quad{ }^{199} X: 8.0 \% \quad{ }^{202} X: 2.0 \%
$$

The weighted average atomic mass of the naturallyoccurring element $X$ is closest to:
(A) 201 amu
(B) 202 amu
(C) 199 amu
(D) 200 amu
[AIPMT 2007]
10. Boron has two stable isotopes, ${ }^{10} \mathrm{~B}(19 \%)$ and ${ }^{11} \mathrm{~B}(81 \%)$. Calculate average atomic weight of boron in the periodic table.
(A) 10.8
(B) 10.2
(C) 11.2
(D) 10.0
[AIPMT 1998]
11. Haemoglobin contains $0.334 \%$ of iron by weight. The molecular weight of haemoglobin is approximately 67200. The number of iron atoms (Atomic weight of Fe is 56) present in one molecule of haemoglobin is:
(A) 4
(B) 6
(C) 3
(D) 2
[AIPMT 1990]

---

### 1.8. Mole Concept and Molar Masses

12. 1.0 g of $\mathrm{H}_{2}$ has same number of molecules as in:
(A) 14 g of $\mathrm{N}_{2}$
(B) 18 g of $\mathrm{H}_{2} \mathrm{O}$
(C) 16 g of CO
(D) 28 g of $\mathrm{N}_{2}$
[Re-NEET 2024]
13. The highest number of helium atoms is in:
(A) 4 u of helium
(B) 4 g of helium
(C) 2.271098 L of helium at STP
(D) 4 mol of helium
[NEET 2024]
14. One mole of carbon atom weights 12 g . The number of atoms in it is equal to:
(Mass of Carbon-12 is $1.9926 \times 10^{-23} \mathrm{~g}$ )
(A) $1.2 \times 10^{23}$
(B) $6.022 \times 10^{22}$
(C) $12 \times 10^{22}$
(D) $6.022 \times 10^{23}$
[NEET Oct. 2020]
15. Which one of the following has maximum number of atoms?
(A) 1 g of $\mathrm{Mg}_{(x)}$ [Atomic mass of $\mathrm{Mg}=24$ ]
(B) 1 g of $\mathrm{O}_{2(g)}$ [Atomic mass of $\mathrm{O}=16$ ]
(C) 1 g of $\mathrm{Li}_{(x)}$ [Atomic mass of $\mathrm{Li}=7$ ]
(D) 1 g of $\mathrm{Ag}_{(x)}$ [Atomic mass of $\mathrm{Ag}=108$ ]
[NEET Sept. 2020]
16. The number of moles of hydrogen molecules required to produce 20 moles of ammonia through Haber's process is:
(A) 20
(B) 30
(C) 40
(D) 10 [NEET National 2019]
17. In which case is the number of molecules of water maximum?
(A) 0.00224 L of water vapours at 1 atm and 273 K
(B) 0.18 g of water
(C) 18 mL of water
(D) $10^{-3} \mathrm{~mol}$ of water
[NEET 2018]
18. When 22.4 L of $\mathrm{H}_{2(g)}$ is mixed with 11.2 L of $\mathrm{Cl}_{2(g)}$, each at STP, the moles of $\mathrm{HCl}_{(g)}$ formed is equal to:
(A) 1 mole of $\mathrm{HCl}_{(g)}$
(B) 2 moles of $\mathrm{HCl}_{(g)}$
(C) 0.5 mole of $\mathrm{HCl}_{(g)}$
(D) 1.5 moles of $\mathrm{HCl}_{(g)}$
[NEET 2016]
19. If Avogadro number $\mathrm{N}_{\mathrm{A}}$, is changed from $6.022 \times 10^{23}$ $\mathrm{mol}^{-1}$ to $6.022 \times 10^{20} \mathrm{~mol}^{-1}$, this would change:
(A) the definition of mass in units of grams
(B) the mass of one mole of carbon
(C) the ratio of chemical species to each other in a balanced equation
(D) the ratio of elements to each other in a compound
[AIPMT Latest July 2015]
20. The number of water molecules is maximum in:
(A) 18 molecules of water
(B) 1.8 g of water
(C) 18 g of water
(D) 18 moles of water
[AIPMT Latest July 2015]
21. A mixture of gases contains $\mathrm{H}_{2}$ and $\mathrm{O}_{2}$ gases in the ratio of $1: 4(\mathrm{w} / \mathrm{w})$. What is the molar ratio of the two gases in the mixture?
(A) $1: 4$
(B) $4: 1$
(C) $16: 1$
(D) $2: 1$
[AIPMT Cancelled 2015]
22. Which has the maximum number of molecules among the following?
(A) $44 \mathrm{~g} \mathrm{CO}_{2}$
(B) $48 \mathrm{~g} \mathrm{O}_{3}$
(C) $8 \mathrm{~g} \mathrm{H}_{2}$
(D) $64 \mathrm{~g} \mathrm{SO}_{2}$
[AIPMT Mains 2011]
23. The number of atoms in 0.1 mol of triatomic gas is: $\left(N_{A}=6.02 \times 10^{23} \mathrm{~mol}^{-1}\right)$
(A) $6.026 \times 10^{22}$
(B) $1.806 \times 10^{23}$
(C) $3.600 \times 10^{23}$
(D) $1.800 \times 10^{22}$
[AIPMT Screening 2010]
24. Volume occupied by one molecule of water (density $=1 \mathrm{~g}$ $\mathrm{cm}^{-3}$ ) is:
(A) $9.0 \times 10^{-23} \mathrm{~cm}^{3}$
(B) $6.023 \times 10^{-23} \mathrm{~cm}^{3}$
(C) $3.0 \times 10^{-23} \mathrm{~cm}^{3}$
(D) $5.5 \times 10^{-23} \mathrm{~cm}^{3}$
[AIPMT Screening 2008]
25. The maximum number of molecules are present in:
(A) 15 L of $\mathrm{H}_{2}$ gas at STP
(B) 5 L of $\mathrm{N}_{2}$ gas at STP
(C) 0.5 g of $\mathrm{H}_{2}$ gas
(D) 10 g of $\mathrm{O}_{2}$ gas
[AIPMT 2004]
26. Which has the maximum molecules?
(A) $7 \mathrm{~g} \mathrm{~N}_{2}$
(B) $2 \mathrm{~g} \mathrm{H}_{2}$
(C) $16 \mathrm{~g} \mathrm{NO}_{2}$
(D) $16 \mathrm{~g} \mathrm{O}_{2}$
[AIPMT 2002]
27. The number of atoms present in 4.25 g of $\mathrm{NH}_{3}$ is approximately:
(A) $4 \times 10^{23}$
(B) $2 \times 10^{23}$
(C) $1 \times 10^{23}$
(D) $6 \times 10^{23}$
[AIPMT 1999]
28. The number of moles of oxygen in one litre of air containing $21 \%$ oxygen by volume, under standard conditions, is:
(A) 0.0093 mol
(B) 2.10 mol
(C) 0.186 mol
(D) 0.21 mol
[AIPMT 1995]
29. The total number of valence electrons in 4.2 g of $\mathrm{N}_{3}^{-}$ion is: $\left(\mathrm{N}_{\mathrm{A}}\right.$ is the Avogadro's number)
(A) $2.1 \mathrm{~N}_{\mathrm{A}}$
(B) $4.2 \mathrm{~N}_{\mathrm{A}}$
(C) $1.6 \mathrm{~N}_{\mathrm{A}}$
(D) $3.2 \mathrm{~N}_{\mathrm{A}}$
[AIPMT 1994]

---

30. The number of gram molecules of oxygen in $6.02 \times 10^{24}$ CO molecules is:
(A) 10 g molecules
(B) 5 g molecules
(C) 1 g molecule
(D) 0.5 g molecule
[AIPMT 1990]
31. Ratio of $C_{p}$ and $C_{V}$ of a gas ' $X$ ' is 1.4 . The number of atoms of the gas ' $X$ ' present in 11.2 litres of it at NTP will be:
(A) $6.02 \times 10^{23}$
(B) $1.2 \times 10^{23}$
(C) $3.01 \times 10^{23}$
(D) $2.01 \times 10^{23}$
[AIPMT 1989]
32. The number of oxygen atoms in 4.4 g of $\mathrm{CO}_{2}$ is:
(A) $1.2 \times 10^{23}$
(B) $6 \times 10^{22}$
(C) $6 \times 10^{23}$
(D) $12 \times 10^{23}$
[AIPMT 1989]
33. At STP, the density of $\mathrm{CCl}_{4}$ vapours in $\mathrm{g} / \mathrm{L}$ will be nearest to:
(A) 6.87
(B) 3.42
(C) 10.26
(D) 4.57
[AIPMT 1988]
34. $1 \mathrm{cc} \mathrm{N}_{2} \mathrm{O}$ at NTP contains:
(A) $\frac{1.8}{224} \times 10^{22}$ atoms
(B) $\frac{6.02}{22400} \times 10^{23}$ molecules
(C) $\frac{1.32}{224} \times 10^{23}$ electrons
(D) All of the above
[AIPMT 1988]

### 1.9. Percentage Composition

35. On complete combustion, 0.3 g of an organic compound gave 0.2 g of $\mathrm{CO}_{2}$ and 0.1 g of $\mathrm{H}_{2} \mathrm{O}$. The percentage composition of carbon and hydrogen in the compound, respectively is:
(A) $4.07 \%$ and $15.02 \%$
(B) $18.18 \%$ and $3.70 \%$
(C) $15.02 \%$ and $4.07 \%$
(D) $3.70 \%$ and $18.18 \%$
[Re-NEET 2024]
36. A compound W contains $32 \%$ of $\mathrm{X}, 20 \%$ of Y and remaining percentage of $Z$. Then, the empirical formula of W is :
(Given atomic masses of $\mathrm{X}=64 ; \mathrm{Y}=40 ; \mathrm{Z}=32 \mathrm{u}$ )
(A) $\mathrm{XYZ}_{3}$
(B) $\mathrm{XY}_{2} \mathrm{Z}_{2}$
(C) $\mathrm{XYZ}_{4}$
(D) $\mathrm{X}_{2} \mathrm{YZ}_{2}$
[NEET 2024]
37. An organic compound contains $78 \%$ (by wt.) carbon and remaining percentage of hydrogen. The right option for the empirical formula of this compound is: (Atomic wt. of C is 12, H is 1 )
(A) CH
(B) $\mathrm{CH}_{2}$
(C) $\mathrm{CH}_{3}$
(D) $\mathrm{CH}_{4}$
[NEET 2021]
38. Magnesium reacts with an element $(X)$ to form an ionic comound. If the ground state electronic configuration of $(X)$ is $1 s^{2} 2 s^{2} 2 p^{3}$, the simplest formula for this compound is:
(A) $\mathrm{Mg}_{2} X$
(B) $\mathrm{Mg} X_{2}$
(C) $\mathrm{Mg}_{2} X_{3}$
(D) $\mathrm{Mg}_{3} X_{2}$
[NEET 2018]
39. An organic compound contains carbon, hydrogen and oxygen. Its elemental analysis gave $\mathrm{C}, 38.71 \%$ and H , $9.67 \%$. The empirical formula of the compound would be:
(A) $\mathrm{CH}_{3} \mathrm{O}$
(B) $\mathrm{CH}_{2} \mathrm{O}$
(C) CHO
(D) $\mathrm{CH}_{4} \mathrm{O}$
[AIPMT Screening 2008]
40. The percentage of $\mathrm{C}, \mathrm{H}$ and N in an organic compound are $40 \%, 13.3 \%$ and $46.7 \%$, respectively, then the empirical formula is:
(A) $\mathrm{C}_{3} \mathrm{H}_{13} \mathrm{~N}_{3}$
(B) $\mathrm{CH}_{2} \mathrm{~N}$
(C) $\mathrm{CH}_{4} \mathrm{~N}$
(D) $\mathrm{CH}_{6} \mathrm{~N}$
[AIPMT 2002]
41. Percentage of Se in peroxidase anhydrous enzyme is $0.5 \%$ by weight (at. wt. $=78.4 \mathrm{~g}$ ), then minimum molecular weight of peroxidase anhydrous enzyme is:
(A) $1.568 \times 10^{4}$
(B) $1.568 \times 10^{3}$
(C) 15.68
(D) $2.136 \times 10^{4}$
[AIPMT 2001]

### 1.10. Stoichiometry and Stoichiometric Calculations

42. The amount of glucose required to prepare 250 mL of $\frac{\mathrm{M}}{20}$ aqueous solution is:
(Molar mass of glucose: $180 \mathrm{~g} \mathrm{~mol}^{-1}$ )
(A) 2.25 g
(B) 4.5 g
(C) 0.44 g
(D) 1.5 g
[Re-NEET 2024]
43. 1 gram of sodium hydroxide was treated with 25 mL of 0.75 M HCl solution, the mass of sodium hydroxide left unreacted is equal to:
(A) 250 mg
(B) Zero mg
(C) 200 mg
(D) 750 mg
[NEET 2024]
44. The right option for the mass of $\mathrm{CO}_{2}$ produced by heating 20 g of $20 \%$ pure limestone is: (Atomic mass of $\mathrm{Ca}=40$ )

$$
\mathrm{CaCO}_{3} \xrightarrow{1200 \mathrm{~K}} \mathrm{CaO}+\mathrm{CO}_{2}
$$

(A) 2.64 g
(B) 1.32 g
(C) 1.12 g
(D) 1.76 g
[NEET 2023]
45. What mass of $95 \%$ pure $\mathrm{CaCO}_{3}$ will be required to neutralize 50 mL of 0.5 M HCl solution according to the following reaction?
$\mathrm{CaCO}_{3(\mathrm{r})}+2 \mathrm{HCl}_{(\mathrm{eq})} \rightarrow \mathrm{CaCl}_{2(\mathrm{eq})}+\mathrm{CO}_{2(\mathrm{~g})}+2 \mathrm{H}_{2} \mathrm{O}_{(\mathrm{l})}$
(Calculate upto second place of decimal point)
(A) 1.25 g
(B) 1.32 g
(C) 3.65 g
(D) 9.50 g
[NEET 2022]

---

46. In one molal solution that contains 0.5 mole of a solute, there is:
(A) 500 mL of solvent
(B) 500 g of solvent
(C) 100 mL of solvent
(D) 1000 g of solvent
[NEET 2022]
47. The density of 2 M aqueous solution of NaOH is $1.28 \mathrm{~g} / \mathrm{cm}^{3}$. The molality of the solution is: (Given that molecular mass of $\mathrm{NaOH}=40 \mathrm{~g} \mathrm{~mol}^{-1}$ )
(A) 1.20 m
(B) 1.56 m
(C) 1.67 m
(D) 1.32 m
[NEET Odisha 2019]
48. A mixture of 2.3 g formic acid and 4.5 g oxalic acid is treated with conc. $\mathrm{H}_{2} \mathrm{SO}_{4}$. The evolved gaseous mixture is passed through KOH pellets. Weight (in g) of the remaining product at STP will be:
(A) 2.8
(B) 3.0
(C) 1.4
(D) 4.4
[NEET 2018]
49. Which of the following is dependent on temperature?
(A) Molality
(B) Molarity
(C) Mole fraction
(D) Weight percentage
[NEET 2017]
50. 20.0 g of a magnesium carbonate sample decomposes on heating to give carbon dioxide and 8.0 g magnesium oxide. What will be the percentage purity of magnesium carbonate in the sample? (Atomic weight of $\mathrm{Mg}=24$ )
(A) 75
(B) 96
(C) 60
(D) 84
[AIPMT Latest July 2015]
51. What is the mass of precipitate formed when 50 mL of $16.9 \%$ solution of $\mathrm{AgNO}_{3}$ is mixed with 50 mL of $5.8 \%$ NaCl solution? ( $\mathrm{Ag}=107.8, \mathrm{~N}=14, \mathrm{O}=16, \mathrm{Na}=23$, $\mathrm{Cl}=35.5$ )
(A) 28 g
(B) 3.5 g
(C) 7 g
(D) 14 g
[AIPMT Latest July 2015]
52. What is the mole fraction of the solute in a 1.00 m aqueous solution?
(A) 0.177
(B) 1.770
(C) 0.0354
(D) 0.0177
[AIPMT Latest July 2015]
53. 1.0 g of magnesium burnt with $0.56 \mathrm{~g} \mathrm{O}_{2}$ in a closed vessel. Which reactant is left in excess and how much?
(At. wt.: $\mathrm{Mg}=24 ; \mathrm{O}=16$ )
(A) $\mathrm{Mg}, 0.16 \mathrm{~g}$
(B) $\mathrm{O}_{2}, 0.16$
(C) $\mathrm{Mg}, 0.44 \mathrm{~g}$
(D) $\mathrm{O}_{2}, 0.28 \mathrm{~g}$
[AIPMT 2014]
54. In an experiment, it is shown that 10 mL of 0.05 M solution of chloride required 10 mL of 0.1 M solution of $\mathrm{AgNO}_{3}$. Which of the following will be the formula of the chloride? ( $X$ stands for the symbol of the element other than chlorine)
(A) $X_{2} \mathrm{Cl}_{2}$
(B) $X \mathrm{Cl}_{2}$
(C) $X \mathrm{Cl}_{4}$
(D) $X_{2} \mathrm{Cl}$
[NEET Karnataka 2013]
55. How many grams of concentrated nitric acid solution should be used to prepare 250 mL of $2.0 \mathrm{M} \mathrm{HNO}_{3}$ ? (The concentrated acid is $70 \% \mathrm{HNO}_{3}$ ).
(A) 45.0 g conc. $\mathrm{HNO}_{3}$
(B) 90.0 g conc. $\mathrm{HNO}_{3}$
(C) 70.0 g conc. $\mathrm{HNO}_{3}$
(D) 54.0 g conc. $\mathrm{HNO}_{3}$
[NEET 2013]
56. $6.02 \times 10^{20}$ molecules of urea are present in 100 mL of its solution. The concentration of solution is:
(A) 0.02 M
(B) 0.01 M
(C) 0.001 M
(D) 0.1 M
[NEET 2013]
57. 25.3 g of sodium carbonate, $\mathrm{Na}_{2} \mathrm{CO}_{3}$ is dissolved in enough water to make 250 mL of solution. If sodium carbonate dissociates completely, molar concentration of sodium ion, $\mathrm{Na}^{+}$and carbonate ion, $\mathrm{CO}_{3}^{-}$are respectively: (Molar mass of $\mathrm{Na}_{2} \mathrm{CO}_{3}=106 \mathrm{~g} \mathrm{~mol}^{-1}$ )
(A) 0.955 M and 1.910 M
(B) 1.910 M and 0.955 M
(C) 1.90 M and 1.910 M
(D) 0.477 M and 0.477 M
[AIPMT Screening 2010]
58. 10 g of hydrogen and 64 g of oxygen were filled in a steel vessel and exploded. Amount of water produced in this reaction will be:
(A) 2 mol
(B) 3 mol
(C) 4 mol
(D) 1 mol
[AIPMT Screening 2009]
59. What volume of oxygen gas $\left(\mathrm{O}_{2}\right)$ measured at $0^{\circ} \mathrm{C}$ and 1 atm is needed to burn completely 1 L of propane gas $\left(\mathrm{C}_{3} \mathrm{H}_{8}\right)$ measured under the same condition?
(A) 7 L
(B) 6 L
(C) 5 L
(D) 10 L
[AIPMT Screening 2008]
60. How many moles of lead (II) chloride will be formed from a reaction between 6.5 g of PbO and 3.2 g of HCl ?
(A) 0.044
(B) 0.333
(C) 0.011
(D) 0.029
[AIPMT Screening 2008]
61. The mass of carbon anode consumed (giving only carbon dioxide) in the production of 270 kg of a aluminium metal from bauxite by the Hall process is: (Atomic mass: $\mathrm{Al}=27$ )
(A) 270 kg
(B) 540 kg
(C) 90 kg
(D) 180 kg
[AIPMT 2005]
62. Assuming fully decomposed, the volume of $\mathrm{CO}_{2}$ released at STP on heating 9.85 g of $\mathrm{BaCO}_{3}$ (At. mass of $\mathrm{Ba}=137$ ) will be:
(A) 1.12 L
(B) 0.84 L
(C) 2.24 L
(D) 4.96 L
[AIPMT 2000]

---

63. In the reaction,

$$
4 \mathrm{NH}_{3(g)}+5 \mathrm{O}_{2(g)} \rightarrow 4 \mathrm{NO}_{(g)}+6 \mathrm{H}_{2} \mathrm{O}_{(l)}
$$

When 1 mole of ammonia and 1 mole of $\mathrm{O}_{2}$ are made to react to completion:
(A) All the oxygen will be consumed.
(B) 1.0 mole of NO will be produced.
(C) 1.0 mole of $\mathrm{H}_{2} \mathrm{O}$ is produced.
(D) All the ammonia will be consumed. [AIPMT 1998]
64. A 5 molar solution of $\mathrm{H}_{2} \mathrm{SO}_{4}$ is diluted from 1 litre to a volume of 10 litres, the normality of the solution will be:
(A) 1 N
(B) 0.1 N
(C) 5 N
(D) 0.5 N
[AIPMT 1991]
65. A metal oxide has the formula $\mathrm{Z}_{2} \mathrm{O}_{3}$. It can be reduced by hydrogen to give free metal and water. 0.1596 g of the metal oxide requires 6 mg of hydrogen for complete reduction. The atomic weight of the metal is:
(A) 27.9
(B) 159.6
(C) 79.8
(D) 55.8
[AIPMT 1989]
66. What is the weight of oxygen required for the complete combustion of 2.8 kg of ethylene?
(A) 2.8 kg
(B) 6.4 kg
(C) 9.6 kg
(D) 96 kg
[AIPMT 1989]
67. One litre hard water contains $12.00 \mathrm{mg} \mathrm{Mg}^{2+}$. Milliequivalents of washing soda required to remove its hardness is:
(A) 1
(B) 12.16
(C) $1 \times 10^{-3}$
(D) $12.16 \times 10^{-3}$ [AIPMT 1988]

# SOLUTIONS 

1. (D) $\mathrm{CH}_{3}-\mathrm{C}-\mathrm{ONa} \quad \frac{\mathrm{NaOH}+\mathrm{CaO}}{\text { Soda Lime }} \mathrm{CH}_{4}+\mathrm{CO}_{2} \uparrow$

Sodium ethanoate
One mole of $\mathrm{CH}_{4}$ weighs $=16 \mathrm{~g}$
Two moles of $\mathrm{CH}_{4}$ will weigh $=32 \mathrm{~g}$
2. (B) Given,
$\mathrm{H}_{2}$ density $=0.089$
Volume $=45 \mathrm{~mL}$ or 0.045 L
Weight $=$ Volume $\times$ density
Weight of $\mathrm{H}_{2}=0.045 \times 0.089$

$$
=0.004005 \mathrm{~g}
$$

Weight of volatile gas $=0.24 \mathrm{~g}$
$\therefore$ Vapour density

$$
\begin{aligned}
& =\frac{\text { Weight of given volume of substance }}{\text { Weight of same volume of hydrogen }} \\
& =\frac{0.24}{0.004005}=59.925=59.93
\end{aligned}
$$

## Related Theory

Volume $=\frac{\text { Mass }}{\text { density }}$
Vapour density $=\frac{\text { Mass of specific volume of substance }}{\text { Mass of same volume of hydrogen }}$

## Caution

- Students should convert volume in Litre from mL before proceeding further.
3.(B) Pressure $=\frac{\text { Force }}{\text { Area }}=\frac{\mathrm{F} \times \mathrm{d}}{\mathrm{A} \times \mathrm{d}}$

$$
\frac{\mathrm{W}}{\mathrm{~V}}=\frac{\text { Energy }}{\text { Volume }}=\frac{\text { Force } \times \text { Density }}{\text { Area } \times \text { Density }}
$$

Hence, pressure will have same dimensions as energy/volume.
4. (D)

| Numbers | $\square$ Significant figures |
| :--: | :--: |
| 161 cm | 3 |
| 0.161 cm | 3 |
| 0.0161 cm | 3 |

## Related Theory

$\rightarrow$ Zeroes between two non-zero digits are significant.
$\rightarrow$ If zero precedes the non-zero digit, than it is considered as non-significant.
5. (C) $\frac{(29.2-20.2)\left(1.79 \times 10^{5}\right)}{1.37}=\frac{9 \times 1.79 \times 10^{5}}{1.37}$

$$
=11.76 \times 10^{5}
$$

Or

$$
=1.18 \times 10^{6}
$$

As the least precise number in the question has 3 significant figures, so we need to round off the answer to three significant figures only.

## Caution

- Students should keep this in mind that answer must be reported in same significant numbers as the least precise value has.

6. (C) According to Avogadro's hypothesis

Vol of gas $(\mathrm{V})=$ no. of moles $(\mathrm{n})$

$$
\begin{aligned}
\text { No. of moles } & =\frac{\text { Weighed mass }}{\text { Molar mass }}=\frac{\mathrm{m}}{\mathrm{M}} \\
\mathrm{~V}_{\mathrm{H}_{2}}: \mathrm{V}_{\mathrm{O}_{2}}: \mathrm{V}_{\mathrm{CH}_{4}} & =\frac{\mathrm{m}_{\mathrm{H}_{2}}}{\mathrm{M}_{\mathrm{H}_{2}}}: \frac{\mathrm{m}_{\mathrm{O}_{2}}}{\mathrm{M}_{\mathrm{O}_{2}}}: \frac{\mathrm{m}_{\mathrm{CH}_{4}}}{\mathrm{M}_{\mathrm{CH}_{4}}} \\
& =\frac{\mathrm{m}_{\mathrm{H}_{2}}}{2}: \frac{\mathrm{m}_{\mathrm{O}_{2}}}{32}: \frac{\mathrm{m}_{\mathrm{CH}_{4}}}{16} \\
& =16: 1: 2
\end{aligned}
$$

---

7. (C) According to Avogadro's Law, "Under the same conditions of temperature and pressure, equal volumes of different gases contain equal number of molecules".
As 1 L of $\mathrm{O}_{2}$ contains $N$ molecules, so 2 L will contain $2 N$ molecules for any gas at same temperature and pressure.
8. (A) Let the atomic masses of $X$ and $Y$ be $\mathrm{A}_{X}$ and $\mathrm{A}_{Y}$

$$
\begin{gathered}
n_{x y_{2}}=0.1=\frac{10}{\mathrm{~A}_{x}+2 \mathrm{~A}_{Y}} \\
\mathrm{~A}_{X}+2 \mathrm{~A}_{Y}=100 \\
n_{X_{3} Y_{2}}=0.05=\frac{9}{3 \mathrm{~A}_{x}+2 \mathrm{~A}_{Y}} \\
3 \mathrm{~A}_{X}+2 \mathrm{~A}_{Y}=180
\end{gathered}
$$

On solving eqs. (i) and (ii), we get

$$
\begin{aligned}
& \mathrm{A}_{X}=40 \mathrm{~g} / \mathrm{mol} \\
& \mathrm{~A}_{Y}=30 \mathrm{~g} / \mathrm{mol}
\end{aligned}
$$

9. (D) Average atomic mass $=\left[\right.$ Atomic mass of $\mathrm{X}_{200} \times$ abundance + Atomic mass of $\mathrm{X}_{199} \times$ abundance + Atomic mass of $\mathrm{X}_{202} \times$ abundance $] / 100$.
Average isotopic mass of

$$
\begin{aligned}
X & =\frac{200 \times 90+199 \times 8+202 \times 2}{90+8+2} \\
& =\frac{18000+1592+404}{100} \\
& =\frac{19996}{100} \\
& =199.96 \mathrm{amu}=200 \mathrm{amu}
\end{aligned}
$$

10. (A) Average atomic weight of Boron

$$
\begin{aligned}
& =\frac{\% \text { of }^{10} \mathrm{~B} \times \text { atomic mass }+\% \text { of }^{11} \mathrm{~B} \times \text { atomic mass. }}{\% \text { of }^{10} \mathrm{~B}+\% \text { of }^{11} \mathrm{~B}} \\
& =\frac{19 \times 10+81 \times 11}{19+81} \\
& =\frac{1081}{100}=10.81 \mathrm{~g}
\end{aligned}
$$

11. (A) Molecular weight of haemoglobin $=67200$

$$
\begin{aligned}
\text { Mass of iron in haemoglobin } & =67200 \times \frac{0.334}{100} \\
& =224.448
\end{aligned}
$$

1 atom of iron has mass $=56$
Then atoms in 224.448 mass $=\frac{224.448}{56}$

$$
\approx 4
$$

12. (A) Same number of moles have same number of molecules.

$$
\text { Moles }=\frac{\text { Weight }}{\text { Mol. wt. }}
$$

Moles of 1.0 g of $\mathrm{H}_{2}=\frac{1}{2}=0.5$
Moles of 14 g of $\mathrm{N}_{2}=\frac{14}{28}=0.5$
Moles of 18 g of $\mathrm{H}_{2} \mathrm{O}=\frac{18}{18}=1$
Moles of 16 g of $\mathrm{CO}=\frac{16}{28}=0.51$
Moles of 28 g of $\mathrm{N}_{2}=\frac{28}{28}=1$
13. (D) $4 u$ helium atom has 4 He atoms.

$$
\begin{aligned}
4 \mathrm{~g} \mathrm{He} & =\frac{4}{4} \text { mole helium } \\
& =1 \times \mathrm{N}_{\mathrm{A}} \text { atoms } \\
2.271098 \mathrm{~L} \mathrm{He} & =\frac{2.271098}{22.71098} \\
& =0.1 \mathrm{~mole} \mathrm{He}=0.1 \times \mathrm{N}_{\mathrm{A}} \text { atoms } \\
4 \mathrm{~mole} \mathrm{He} & =4 \times \mathrm{N}_{\mathrm{A}} \mathrm{He} \text { atoms }
\end{aligned}
$$

So, highest number of helium atoms are present in 4 mole of helium.
14. (D) The relation between the no. of atoms and molar mass can be shown as below:
1 Mole $=$ Molar mass $=6.022 \times 10^{23}$ atoms
1 Mole of carbon $=12 \mathrm{~g}=6.022 \times 10^{23}$ atoms of carbon.
15. (C) No. of atoms $=\frac{\text { Given mass }}{\text { Molar mass }} \times \mathrm{N}_{A}$

$$
\begin{aligned}
& \text { For (A); } \\
& \mathrm{Mg}=\frac{1}{24} \times \mathrm{N}_{\mathrm{A}} \\
& \text { For (B); } \\
& \mathrm{O}_{2}=\frac{2}{32} \times \mathrm{N}_{\mathrm{A}}=\frac{1}{16} \times \mathrm{N}_{\mathrm{A}} \\
& \text { For (C); } \\
& \mathrm{Li}=\frac{1}{7} \times \mathrm{N}_{\mathrm{A}} \\
& \text { For (D); } \\
& \mathrm{Ag}=\frac{1}{108} \times \mathrm{N}_{\mathrm{A}}
\end{aligned}
$$

Thus, 1 g of Li has maximum no. of atoms.
16. (B) $\mathrm{N}_{2(g)}+3 \mathrm{H}_{2(g)} \rightleftharpoons 2 \mathrm{NH}_{3(g)}$

2 moles of $\mathrm{NH}_{3}$ require 3 moles of $\mathrm{H}_{2}$
$\therefore 1$ mole of $\mathrm{NH}_{3}$ will require $=\frac{3}{2}$ moles of $\mathrm{H}_{2}$
$\therefore 20$ moles of $\mathrm{NH}_{3}$ will require $=\frac{3}{2} \times 20$ moles of $\mathrm{H}_{2}$
$=30$ moles of $\mathrm{H}_{2}$

---

## Caution

Students should not get confused about limiting reagent as there is no mention of any limiting reagent in the reaction.
17. (C) No. of molecules $=$ no. of moles $\times \mathrm{N}_{\mathrm{A}}$

$$
=\frac{18}{18} \times 6.022 \times 10^{23}
$$

For water, volume $=$ Mass (as density $=1 \mathrm{~g} / \mathrm{mL}$ ) $=6.022 \times 10^{23}$ molecules of $\mathrm{H}_{2} \mathrm{O}$.
18. (A) At STP, 1 mole of any gas occupies 22.4 L of volume.

$$
\mathrm{H}_{2}+\mathrm{Cl}_{2} \longrightarrow 2 \mathrm{HCl}
$$

22.4 L 11.2 L

Here, $\mathrm{Cl}_{2}$ is limiting reagent.
1 mole of $\mathrm{Cl}_{2} \Rightarrow 2$ moles of HCl
$\therefore 0.5$ mole of $\mathrm{Cl}_{2}$ will form 1 mole of HCl .
19. (B) 1 mole of carbon has mass $=12 \mathrm{~g}$ or $6.022 \times 10^{23}$ atoms of carbon have mass $=12 \mathrm{~g}$ $\therefore 6.022 \times 10^{20}$ atoms of carbon will have mass

$$
\begin{aligned}
& =\frac{12}{6.022 \times 10^{23}} \times 6.022 \times 10^{20} \\
& =0.012 \mathrm{~g}
\end{aligned}
$$

20. (D) 18 moles of water contain $=18 \times 6.022 \times 10^{23}$

$$
=108.396 \times 10^{23} \text { molecules. }
$$

## Related Theory

- Molar mass $=6.022 \times 10^{23}$ atoms/molecules/ions
No. of moles $=\frac{\text { Weighted mass }}{\text { Molar mass }}$
- Mole is the biggest unit to measure number of molecules/atoms/ ions.

21. (B) Molar ratio $=\frac{n_{\mathrm{H}_{2}}}{n_{\mathrm{O}_{2}}}=\frac{x / 2}{4 x / 32}$

$$
\begin{aligned}
& =\frac{x \times 32}{2 \times 4 x} \\
& =\frac{4}{1}=4: 1
\end{aligned}
$$

22. (C) Let us calculate the no. of molecules in each of the given options:

$$
\begin{aligned}
44 \mathrm{~g} \mathrm{CO}_{2} & =1 \mathrm{~mol} \mathrm{CO}_{2}=1 \times \mathrm{N}_{\mathrm{A}} \\
& =6.022 \times 10^{23} \text { molecules } \mathrm{CO}_{2} \\
48 \mathrm{~g} \text { of } \mathrm{O}_{3} & =1 \mathrm{~mol} \mathrm{O}_{3}=1 \times \mathrm{N}_{\mathrm{A}} \\
& =6.022 \times 10^{23} \text { molecules of } \mathrm{O}_{3} \\
8 \mathrm{~g} \text { of } \mathrm{H}_{2} & =4 \text { moles } \mathrm{H}_{2}=4 \times \mathrm{N}_{\mathrm{A}} \\
& =24.088 \times 10^{23} \text { molecules of } \mathrm{H}_{2} \\
64 \mathrm{~g} \mathrm{SO}_{2} & =1 \mathrm{~mol} \mathrm{SO}_{2}=1 \times \mathrm{N}_{\mathrm{A}} \\
& =6.022 \times 10^{23} \text { molecules of } \mathrm{SO}_{2}
\end{aligned}
$$

## Related Theory

- I mole of any substance $=6.022 \times 10^{23}$ particles (atoms/ molecules ions).

23. (B) No. of atoms $=$ No. of moles $\times \mathrm{N}_{\mathrm{A}} \times$ Atomicity

$$
\begin{aligned}
& =0.1 \times 6.02 \times 10^{23} \times 3 \\
& =1.806 \times 10^{23} \text { atoms }
\end{aligned}
$$

24. (C) Weight of $6.023 \times 10^{23}$ molecules of water $=18 \mathrm{~g}$ Volume occupied by $6.023 \times 10^{23}$ molecules of water (density $=1 \mathrm{~g} \mathrm{~cm}^{-3}$ ) will be:

$$
=\frac{18 \mathrm{~g}}{\lg \mathrm{~cm}^{-3}}=18 \mathrm{~cm}^{3} \text { or } \mathrm{mL}
$$

So, volume occupied by one molecule of water

$$
\begin{aligned}
& =\frac{18}{6.023 \times 10^{23}}=2.988 \times 10^{-23} \mathrm{~cm}^{3} \\
& =3.0 \times 10^{-23} \mathrm{~cm}^{3}
\end{aligned}
$$

25. (A) At STP, $22.4 \mathrm{~L} \mathrm{H}_{2}$ has $6.023 \times 10^{23}$ molecules

$$
\begin{aligned}
15 \mathrm{~L} \text { of } \mathrm{H}_{2} & =\frac{6.023 \times 10^{23} \times 15}{22.4} \\
& =4.033 \times 10^{23} \text { molecules } \\
5 \mathrm{~L} \text { of } \mathrm{N}_{2} & =\frac{6.023 \times 10^{23} \times 5}{22.4} \\
& =1.344 \times 10^{23} \text { molecules } \\
0.5 \mathrm{~g} \text { of } \mathrm{H}_{2} & =\frac{6.023 \times 10^{23} \times 0.5}{2} \\
& =1.505 \times 10^{23} \text { molecules } \\
10 \mathrm{~g} \text { of } \mathrm{O}_{2} & =\frac{6.023 \times 10^{23} \times 10}{32} \\
& =1.882 \times 10^{23} \text { molecules }
\end{aligned}
$$

Hence, the maximum no. of molecules are present in 15 L of $\mathrm{H}_{2}$.
26. (B) No. of moles $=\frac{\text { Given mass }}{\text { Molecular weight }}$

$$
\begin{aligned}
7 \mathrm{~g} \mathrm{~N}_{2} & \rightarrow \frac{7}{28}=0.25 \mathrm{~mole} \\
2 \mathrm{~g} \mathrm{H}_{2} & \rightarrow \frac{2}{2}=1 \mathrm{~mole} \\
16 \mathrm{~g} \mathrm{NO}_{2} & \rightarrow \frac{16}{46}=0.347 \mathrm{~mole} \\
16 \mathrm{~g} \mathrm{O}_{2} & \rightarrow \frac{16}{32}=0.5 \mathrm{~mole}
\end{aligned}
$$

Number of moles $\propto$ Number of molecules.
Hence, $2 \mathrm{~g} \mathrm{H}_{2}$ has maximum molecules.
27. (D) No. of moles of $\mathrm{NH}_{3}$

$$
\begin{aligned}
& =\frac{\text { Given weight of } \mathrm{NH}_{3}}{\text { Molecular weight of } \mathrm{NH}_{3}} \\
& =\frac{4.25}{17}=0.25 \mathrm{~mole}
\end{aligned}
$$

1 mole of $\mathrm{NH}_{3}$ contains $6.022 \times 10^{23}$ molecules 0.25 mole of $\mathrm{NH}_{3}$ contains

$$
\begin{aligned}
& =\frac{6.022 \times 10^{23}}{1} \times 0.25 \\
& =1.50 \times 10^{23} \text { molecules }
\end{aligned}
$$

---

1 molecule of $\mathrm{NH}_{3}$ contains 4 atoms;
$\because 1.50 \times 10^{23}$ molecules contains

$$
\begin{aligned}
& =1.50 \times 10^{23} \times 4 \\
& =6.0 \times 10^{23} \text { atoms }
\end{aligned}
$$

28. (A)

Volume of oxygen $=\frac{21}{100} \times 1$

$$
=0.21 \text { litres }=210 \mathrm{~cm}^{3}
$$

At STP, 1 mole oxygen $=22400 \mathrm{~cm}^{3}$
Therefore, moles of oxygen $=\frac{210}{22400}$ moles

$$
=0.0093
$$

Therefore, number of moles of oxygen are 0.0093 moles.

## Related Theory

- According to the mole concept, I mol of a gas at STP has a volume of 22400 mL or 22.4 L .


## $\triangle$ Caution

- Always take same units for volume. For eg., if volume, is taken in $m L$ then I mol will have 22400 mL not 22.4 mL .
29. (C) Number of moles of $\mathrm{N}_{3}^{-}=\frac{4.2}{42}=0.1$

1 mole contains $\mathrm{N}_{\mathrm{A}}$ atoms
So, 0.1 mole will contain $0.1 \mathrm{~N}_{\mathrm{A}}$ atoms
1 atom of $\mathrm{N}_{3}^{-}$has $5+5+5+1$

$$
=16 \text { valence electrons }
$$

So, $\quad 0.1 \mathrm{~N}_{\mathrm{A}}$ atoms $=0.1 \times 16 \times \mathrm{N}_{\mathrm{A}}$

$$
=1.6 \mathrm{~N}_{\mathrm{A}} \text { valence electrons }
$$

## $\triangle$ Caution

- Students usually get confused about valence electrons. They should understand that valence electrons are the electrons present in outermost shell. Therefore, $N$ has 5 valence electrons, not 7 electrons.

30. (B) Moles of CO molecules $=\frac{6.02 \times 10^{24}}{6.02 \times 10^{23}}$

$$
=10 \text { moles }
$$

Mass of oxygen in 10 moles $=160 \mathrm{~g}$

$$
\text { g-molecules of } \mathrm{O}_{2}=\frac{160}{32}=5 \text { g-molecules. }
$$

## Related Theory

- A gram molecule is defined as the number of grams of a molecule which is same as the relative molecular mass of that particular molecule. Like one g-molecule of $\mathrm{O}_{2}$ has 32 g $(16+16)$.
31. (A) $\frac{C_{P}}{C_{V}}=1.4$ (Represents diatomic gas)

Moles of given gas $=\frac{\text { Volume given (L) }}{22.4}$

$$
=\frac{11.2}{22.4}=0.5
$$

0.5 moles have $0.5 \times \mathrm{N}_{\mathrm{A}}$ molecules

As the gas is diatomic, so number of atoms

$$
\begin{aligned}
& =0.5 \times \mathrm{N}_{\mathrm{A}} \times 2=\mathrm{N}_{\mathrm{A}} \text { atoms } \\
& =6.023 \times 10^{23}
\end{aligned}
$$

32. (A)

$$
4.4 \mathrm{~g} \text { of } \mathrm{CO}_{2}=\frac{4.4}{44}=0.1 \mathrm{~mole}
$$

$$
=0.1 \mathrm{~N}_{\mathrm{A}} \text { molecules }
$$

1 molecule of $\mathrm{CO}_{2}$ contains 2 oxygen atoms
Hence, $0.1 \mathrm{~N}_{\mathrm{A}}$ molecules of $\mathrm{CO}_{2}$

$$
\begin{aligned}
& =0.1 \times \mathrm{N}_{\mathrm{A}} \times 2 \text { oxygen atoms } \\
& =0.2 \times 6.023 \times 10^{23} \\
& \approx 1.2 \times 10^{23}
\end{aligned}
$$

33. (A) Mass of $\mathrm{CCl}_{4}=12+4 \times 35.5=154 \mathrm{~g}$

Volume at STP for any gas $=22.4 \mathrm{~L}$

$$
\text { Density }=\frac{\text { Mass }}{\text { Volume }}=\frac{154}{22.4}=6.875 \mathrm{~g} / \mathrm{L}
$$

34. (D) At STP volume of $1 \mathrm{~mol}=22.4 \mathrm{~L}$

$$
\begin{aligned}
& =22400 \mathrm{cc} \\
& \left(\because 1 \mathrm{~L}=1000 \mathrm{cc} \text { or } \mathrm{cm}^{3}\right) \\
\text { Moles of } 1 \mathrm{cc} & =\frac{1}{22400} \\
\text { Molecules in } 1 \mathrm{cc} & =\frac{1}{22400} \times \mathrm{N}_{\mathrm{A}} \\
& =\frac{1}{22400} \times 6.023 \times 10^{23}
\end{aligned}
$$

As 1 molecule of $\mathrm{N}_{2} \mathrm{O}$ contains 3 atoms
So, atoms in $\frac{6.023}{22400} \times 10^{23}$ molecules

$$
\begin{aligned}
& =\frac{6.023}{22400} \times 10^{23} \times 3 \\
& =\frac{1.8}{224} \times 10^{22}
\end{aligned}
$$

Electrons in 1 molecule of $\mathrm{N}_{2} \mathrm{O}$ are $(7+7+8)=22$
So, electrons in given molecule

$$
\begin{aligned}
& =\frac{22 \times 6.023 \times 10^{23}}{22400} \\
& =\frac{1.32}{224} \times 10^{23}
\end{aligned}
$$

---

## $\triangle$

Caution

- Students should always take care of units. 1 mole has 22400 $\mathrm{cm}^{3}$ or 22.4 L volume not $22.4 \mathrm{~cm}^{3}$.
- Centimetre cube is called cc.

35. (B) $\%$ of $\mathrm{H}=\frac{2}{18} \times \frac{x}{w} \times 100 \%$
$\%$ of $\mathrm{C}=\frac{12}{44} \times \frac{y}{w} \times 100 \%$
$x=$ amount of water produced
$y=$ amount of carbon dioxide produced
$w=$ amount of organic compound
Given, $x=0.1 \mathrm{~g}, y=0.2 \mathrm{~g}$ and $w=0.3 \mathrm{~g}$
Putting the values we get
$\%$ of $\mathrm{H}=\frac{2}{18} \times \frac{0.1}{0.3} \times 100 \%=3.7 \%$
$\%$ of $\mathrm{C}=\frac{12}{44} \times \frac{0.2}{0.3} \times 100=18.18 \%$
36. (A)

| Element | $\%$ | Mole | Molar <br> ratio |
| :--: | :--: | :--: | :--: |
| X | 32 | $\frac{32}{64}=\frac{1}{2}$ | 1 |
| Y | 20 | $\frac{20}{40}=\frac{1}{2}$ | 1 |
| Z | 48 | $\frac{48}{32}=\frac{3}{2}$ | 3 |

Empirical Formula $=\mathrm{XYZ}_{3}$
37. (C)

| Element | $\%$ | At. <br> weight | Moles | Simplest <br> ratio |
| :--: | :--: | :--: | :--: | :--: |
| C | 78 | 12 | 6.5 | 1 |
| H | 22 | 1 | 22 | 3 |

Hence, empirical formula of this compound is $\mathrm{CH}_{3}$.
38. (D) $\mathrm{X}=1 s^{2} 2 s^{2} 2 p^{3}$

Valency of $\mathrm{X}=-3$
Valency of $\mathrm{Mg}=+2$
By cross multiplying,
![img-0.jpeg](img-0.jpeg)

The formula is $\mathrm{Mg}_{3} \mathrm{X}_{2}$
39. (A) Atomic mass of $\mathrm{C}=12, \mathrm{H}=1$ and $\mathrm{O}=16$

| Element | Percentage | Molar <br> ratio | Simplest <br> molar ratio |
| :--: | :--: | :--: | :--: |
| C | 38.71 | 3.22 | 1 |
| H | 9.67 | 9.67 | 3 |
| O | 51.62 | 3.22 | 1 |

Thus, empirical formula of the compound is $\mathrm{CH}_{3} \mathrm{O}$.
40. (C)

| Ele- <br> ment | $\%$ | At. Wt. | Number of <br> moles | Simplest <br> whole number <br> ratio |
| :--: | :--: | :--: | :--: | :--: |
| C | 40.0 | 12 | $\frac{40}{12}=3.33$ | $\frac{3.33}{3.33}=1$ |
| H | 13.3 | 1 | $\frac{13.3}{1}=13.3$ | $\frac{13.3}{3.33}=4$ |
| N | 46.7 | 14 | $\frac{46.7}{14}=3.33$ | $\frac{3.33}{3.33}=1$ |

Hence, empirical formula of compound is $\mathrm{CH}_{4} \mathrm{~N}$.
41. (A) $0.5 \%$ of Se by weight means 0.5 g of Se is present in 100 g of peroxidase anhydrous enzyme, i.e., 0.5 g of Se is present in 100 g of enzyme.

$$
\begin{aligned}
\therefore 78.4 \mathrm{~g} \text { of } \mathrm{Se} \text { is present in } \frac{100}{0.5} \times 78.4 & =15680 \mathrm{~g} \\
& =1.568 \times 10^{4} \mathrm{~g}
\end{aligned}
$$

42. (A)

$$
\text { Molarity }=\frac{\text { Moles of solute } \times 1000}{\text { Volume of solution }(\mathrm{mL})}
$$

$$
\begin{aligned}
\mathrm{V}(\mathrm{~mL}) & =250 \mathrm{~mL} \\
\text { Molarity } & =\frac{\mathrm{M}}{20}
\end{aligned}
$$

Thus $\quad \frac{1}{20}=\frac{\text { Moles of glucose } \times 1000}{250}$
Moles of glucose $=\frac{1}{80}$
Moles of glucose $=\frac{\text { Mass of glucose }}{\text { Molar mass }}$
Mass of glucose $=$ Moles $\times$ Molar mass

$$
=\frac{1}{80} \times 180=2.25 \mathrm{~g}
$$

43. (A) Balanced reaction can be shown as:

$$
\begin{aligned}
& \mathrm{NaOH}+\mathrm{HCl} \longrightarrow \mathrm{NaCl}+\mathrm{H}_{2} \mathrm{O} \\
& 1 \text { mole } 1 \text { mole } \\
& 40 \mathrm{~g} \quad 36.5 \mathrm{~g} \\
& \text { Mole of } \mathrm{HCl}=\frac{\text { Molarity } \times \text { Volume }(\mathrm{mL})}{1000}
\end{aligned}
$$

---

$$
=\frac{0.75 \times 25}{1000}
$$

As,

$$
\begin{aligned}
\text { mole } & =\frac{\text { weight }}{\text { mol. weight }} \\
\text { weight } & =\text { mole } \times \text { molecular weight } \\
\text { mass of } \mathrm{HCl} & =\frac{0.75 \times 25 \times 36.5}{1000}
\end{aligned}
$$

36.5 g HCl react with $40 \mathrm{~g} \mathrm{NaOH} \frac{0.75 \times 25 \times 36.5}{1000} \mathrm{~g}$

HCl would react with $\frac{40}{36.5} \times \frac{0.75 \times 25 \times 36.5}{1000}=0.75 \mathrm{~g}$
$\mathrm{NaOH}$ left unreacted $=1-0.75 \mathrm{~g}$

$$
\begin{aligned}
& =0.25 \mathrm{~g} \\
& =250 \mathrm{mg}
\end{aligned}
$$

44. (D) The balanced equation is given as:

| $\mathrm{CaCO}_{3}$ | $\xrightarrow{1200 \mathrm{~K}}$ | $\mathrm{CaO}+\mathrm{CO}_{2}$ |
| :--: | :--: | :--: |
| 1 mol |  | 1 mol 1 mol |
| $(100 \mathrm{~g})$ |  | $(56 \mathrm{~g})$ |
| 100 g of $\mathrm{CaCO}_{3}$ produces 44 g of $\mathrm{CO}_{2}$ |  |  |
| 20 g of $\mathrm{CaCO}_{3}$ will produce |  |  |
|  |  |  |

Thus 8.8 of $\mathrm{CO}_{2}$ will be released if $\mathrm{CaCO}_{3}$ is $100 \%$ pure but purity of sample is only $20 \%$. Thus, $\mathrm{CO}_{2}$ formed from $20 \mathrm{~g} \mathrm{CaCO}_{3}$ will be

$$
=8.8 \times \frac{20}{100}=1.76 \mathrm{~g}
$$

45. (B) The given equation is
$\mathrm{CaCO}_{3(\mathrm{r})}+2 \mathrm{HCl}_{(\mathrm{aq})} \rightarrow \mathrm{CaCl}_{2(\mathrm{aq})}+\mathrm{CO}_{2(g)}+\mathrm{H}_{2} \mathrm{O}_{(l)}$
That shows 1 mole of $\mathrm{CaCO}_{3}$ is required for 2 moles of HCl .
Millimoles of HCl = Molarity $\times$ Volume (in mL )

$$
=0.5 \times 50=25
$$

Millimoles of $\mathrm{CaCO}_{3}$ required $=\frac{25}{2}$

$$
\begin{aligned}
& =12.5 \\
\text { Mass of } 12.5 \mathrm{mmol} \text { of } \mathrm{CaCO}_{3} & =12.5 \times 100 \mathrm{mg} \\
& =1.250 \mathrm{~g}
\end{aligned}
$$

If the sample is $95 \%$ pure, then mass of $\mathrm{CaCO}_{3}$ required $=1.250 \times \frac{100}{95}=1.32 \mathrm{~g}$

## A) Caution

- Students in hurry forget that Molarity $\times$ Volume in mL gives number of millimoles, i.e. $10^{-3}$ moles not 1 mole.
- Students should first do calculation considering $100 \%$ purity, then must change the answer according to the given purity.

46. (B) Mass of solvent $=\frac{\text { Moles of solute }}{\text { Molality }}$

$$
\begin{aligned}
& =\frac{0.5}{1} \\
& =0.5 \mathrm{~kg} \text { or } 500 \mathrm{~g}
\end{aligned}
$$

## Related Theory

Molality $=\frac{\text { Moles of solute }}{\text { Mass of solvent }(\mathrm{kg})}$
47. (C) 2 M solution of NaOH means 2 moles of NaOH are present in 1 L solution

$$
\text { Density }=1.28 \mathrm{~g} / \mathrm{mL}
$$

Mass of solution $=$ Volume of solution $\times$ Density

$$
\begin{aligned}
& =1000 \times 1.28 \\
& =1280 \mathrm{~g}
\end{aligned}
$$

No. of moles of solute $=\frac{\text { Mass of solute }}{\text { Molar mass of solution }}$

$$
2=\frac{\text { Mass of solute }}{40 \mathrm{~g} / \mathrm{mol}}
$$

Mass of solute $=80 \mathrm{~g} / \mathrm{mol}$
Mass of solvent $=$ Mass of solution

$$
\begin{aligned}
& =1280-80 \\
& =1200 \mathrm{~g}=1.2 \mathrm{~kg}
\end{aligned}
$$

Molality $=\frac{\text { Moles of solute }}{\text { Mass of solvent }}$
Molality of solution $=\frac{2}{1.2}$

$$
=1.67 \mathrm{~m}
$$

48. (A) Reaction 1:

$$
\begin{aligned}
& \mathrm{HCOOH} \xrightarrow{\text { conc. } \mathrm{H}_{2} \mathrm{SO}_{4}} \mathrm{CO}_{(g)}+\mathrm{H}_{2} \mathrm{O}_{(l)} \\
& \text { Initial } \\
& \frac{2.3}{46} \sim \frac{1}{20} \mathrm{~mol} \quad 0 \\
& \text { Final } \\
& \text { moles } \quad 0 \quad \frac{1}{20}
\end{aligned}
$$

Reaction 2:

$$
\begin{aligned}
& \mathrm{COOH} \xrightarrow{\text { conc. } \mathrm{H}_{2} \mathrm{SO}_{4}} \mathrm{CO}_{(g)}+\mathrm{CO}_{2(g)}+\mathrm{H}_{2} \mathrm{O}_{(l)} \\
& \text { Initial } \\
& \text { moles } \frac{4.5}{90} \sim \frac{1}{20} \mathrm{~mol} \quad 0 \quad 0 \\
& \text { Final } \\
& \text { moles } \quad 0 \quad \frac{1}{20} \quad \frac{1}{20}
\end{aligned}
$$

As the gaseous mixture ( CO and $\mathrm{CO}_{2}$ ) is passed through KOH pellets, hence $\mathrm{CO}_{2(g)}$ will be absorbed by KOH . Water is absorbed by $\mathrm{H}_{2} \mathrm{SO}_{4}$. Hence, remaining product will be $\mathrm{CO}_{(g)}$ only.

---

Total no. of moles of CO formed in above reactions

$$
=\frac{1}{20}+\frac{1}{20}=\frac{1}{10}
$$

Weight of the remaining product

$$
\begin{aligned}
& =\text { No. of moles } \times \text { Molar mass } \\
& =\frac{1}{10} \times 28=2.8 \mathrm{~g}
\end{aligned}
$$

## Related Theory

- I mole $=$ Molar mass $=$ Avogadro's number.

49. (B)

$$
\text { Molality }=\frac{\text { Moles of solute }}{\text { Mass of solvent }(\mathrm{kg})}
$$

Molality does not depend on the temperature but it depends on the mass of solute and solvent.

$$
\text { Molarity }=\frac{\text { Moles of solute }}{\text { Volume of solution (L) }}
$$

As volume depends on temperature hence, molarity is temperature dependent.
50. (D) $\mathrm{MgCO}_{3(\alpha)} \longrightarrow \mathrm{MgO}_{(\alpha)}+\mathrm{CO}_{2(g)}$

Moles of $\mathrm{MgCO}_{3}=\frac{20}{84}=0.238 \mathrm{~mol}$
$\therefore 0.238$ mole $\mathrm{MgCO}_{3}$ will give $=0.238 \mathrm{~mol}$ of MgO

$$
=0.238 \times 40=9.52 \mathrm{~g} \mathrm{MgO}
$$

Practical yield of $\mathrm{MgO}=8 \mathrm{~g}$

$$
\begin{aligned}
& \% \text { purity }=\frac{8}{9.52} \times 100 \\
& =84 \%
\end{aligned}
$$

51. (C) $\mathrm{AgNO}_{3}+\mathrm{NaCl} \longrightarrow \mathrm{AgCl}+\mathrm{NaNO}_{3}$

$$
\begin{aligned}
& \begin{array}{l}
\text { Initial } \\
\text { moles }
\end{array} \frac{8.45}{169.8} \quad \begin{array}{l}
2.9 \\
58.5 \\
=0.049
\end{array} \quad 0 \quad 0 \\
& \text { After } \\
& \text { reaction } 0 \quad 0 \quad 0.049 \quad 0.049 \\
& \text { Mass of } \mathrm{AgCl} \text { ppt }=0.049 \times 143.5 \\
& =7.03 \mathrm{~g}
\end{aligned}
$$

52. (D)

Molality $=\frac{\text { Moles of solute }}{\text { Mass of solvent }(\mathrm{kg})}$
1 Molal solution means 1 mole of solute dissolve in 1000 g of solvent.

Moles of solvent $=\frac{1000}{18}=55.55$
Moles of reaction (Mole fraction of solute)

$$
\begin{aligned}
& =\frac{\text { Moles of solute }}{\text { Total moles }} \\
& =\frac{1}{1+55.55}=0.0177
\end{aligned}
$$

53. (A) $\mathrm{Mg}+\frac{1}{2} \mathrm{O}_{2} \rightarrow \mathrm{MgO}$
$24 \mathrm{~g} \quad 16 \mathrm{~g} \quad 40 \mathrm{~g}$
24 g of Mg reacts with 16 g of $\mathrm{O}_{2}$
Then 1 g of Mg will react with

$$
\frac{16}{24} \times 1=0.67 \mathrm{~g} \mathrm{O}_{2}
$$

However, only 0.56 g of $\mathrm{O}_{2}$ is available. Thus, $\mathrm{O}_{2}$ is a limiting reagent. Now, 0.56 g of $\mathrm{O}_{2}$ reacts with Mg

$$
=\frac{24}{16} \times 0.56=0.84 \mathrm{~g}
$$

Amount of Mg left unreacted $=1.0-0.84$

$$
=0.16 \mathrm{~g}
$$

54. (B) Molarity $=\frac{\text { No. of moles of chloride }}{\text { Volume (in Lt) }}$

$$
0.05=\frac{\text { No. of moles of chloride } \times 1000}{10}
$$

No of moles of chloride $=\frac{0.05 \times 10}{1000}=\frac{0.5}{1000}$
gram equivalent of chloride $=\frac{0.5 \mathrm{n}}{1000}$

$$
\text { No. of moles of } \mathrm{AgNO}_{3}=\frac{0.1 \times 10}{1000}=\frac{1}{1000}
$$

For $\mathrm{AgNO}_{3}, \quad \mathrm{n}$ factor $=1$
gram equivalent of chloride $=$ gram equivalent of $\mathrm{AgNO}_{3}$

$$
\begin{aligned}
\frac{0.5 \mathrm{n}}{1000} & =\frac{1}{1000} \\
\mathrm{n} & =\frac{1}{0.5} \\
\mathrm{n} & =2
\end{aligned}
$$

Therefore, the formula will be $\mathrm{XCl}_{2}$.
55. (A)

$$
\text { Molarity }=\frac{\text { Moles of } \mathrm{HNO}_{3}}{\text { Volume of solution in } \mathrm{L}}
$$

Moles of $\mathrm{HNO}_{3}=2 \times \frac{250}{1000}=0.5$

$$
\begin{aligned}
\text { Mass of } \mathrm{HNO}_{3} & =\text { Moles } \times \text { Molar mass } \\
& =0.5 \times 63=31.5 \mathrm{~g} \mathrm{HNO}_{3}
\end{aligned}
$$

If acid is $100 \%$ mass of $\mathrm{HNO}_{3}=31.5 \mathrm{~g}$
If acid is $70 \%$ mass of $\mathrm{HNO}_{3}$

$$
=31.5 \times \frac{100}{70}=45 \mathrm{~g}
$$

---

56. (B) No. of moles of urea $=\frac{6.02 \times 10^{20}}{6.022 \times 10^{23}}$

$$
=1 \times 10^{-3} \mathrm{~mol}
$$

Volume of solution $=100 \mathrm{~mL}=0.1 \mathrm{~L}$
Molarity of urea solution $=\frac{1 \times 10^{-3}}{0.1}=0.01 \mathrm{M}$
57. (B) Molarity of solution $=\frac{\text { No. of moles of solute }}{\text { Vol. of solution in } \mathrm{L}}$

$$
=\frac{25.3 \times 1000}{106 \times 250}
$$

$$
=0.955 \mathrm{M}
$$

$\mathrm{Na}_{2} \mathrm{CO}_{3}$ dissociates as:
$\mathrm{Na}_{2} \mathrm{CO}_{3} \rightleftharpoons 2 \mathrm{Na}^{+}+\mathrm{CO}_{3}^{2-}$
$x$
$2 x$
$x$
$\therefore \quad$ Conc. of $\mathrm{Na}^{+}=2 \times 0.955$
$=1.910 \mathrm{M}$
$\therefore \quad$ Conc. of $\mathrm{CO}_{3}^{2-}=0.955 \mathrm{M}$
So, concentration of $\mathrm{Na}^{+}$and $\mathrm{CO}_{3}^{2-}$ are 1.910 M and 0.955 M respectively
58. (C) The balanced equation can be written as:

$$
\begin{aligned}
& \mathrm{H}_{2}+\frac{1}{2} \mathrm{O}_{2} \rightarrow \mathrm{H}_{2} \mathrm{O} \\
& 1 \mathrm{~mol} \quad 0.5 \mathrm{~mol} \quad 1 \mathrm{~mol} \\
& \text { Moles of } \mathrm{O}_{2}=\frac{64}{32}=2 \text { moles } \\
& \text { Moles of } \mathrm{H}_{2}=\frac{10}{2}=5 \text { moles }
\end{aligned}
$$

As 1 mole of $\mathrm{H}_{2}$ needed 0.5 mole of $\mathrm{O}_{2}$.
Hence 5 moles of $\mathrm{H}_{2}$ will require 2.5 moles of $\mathrm{O}_{2}$.
Thus, $\mathrm{O}_{2}$ is a limiting reagent. Now, $\mathrm{O}_{2}$ will decide the amount of product formed.
$\frac{1}{2} \mathrm{~mol} \mathrm{O}_{2}$ gives 1 mol of $\mathrm{H}_{2} \mathrm{O}$
2 mole $\mathrm{O}_{2}$ will give 4 moles of $\mathrm{H}_{2} \mathrm{O}$
59. (C) $\mathrm{C}_{3} \mathrm{H}_{8(g)}+5 \mathrm{O}_{2(g)} \longrightarrow 3 \mathrm{CO}_{2(g)}+4 \mathrm{H}_{2} \mathrm{O}_{(l)}$

$$
\begin{aligned}
& 22.4 \mathrm{~L} \text { at STP } 5 \times 22.4 \mathrm{~L} \text { at STP } \\
& \because 22.4 \mathrm{~L} \mathrm{C}_{3} \mathrm{H}_{8} \text { at STP reacts with }=5 \times 22.4 \mathrm{~L} \text { of } \mathrm{O}_{2} \\
& \text { at STP } \\
& \therefore 1 \mathrm{~L} \mathrm{C}_{3} \mathrm{H}_{8} \text { at STP will react with } \frac{5 \times 22.4}{22.4} \text { of } \mathrm{O}_{2} \text { at } \\
& \text { STP } \\
& =5 \mathrm{~L} \text { of } \mathrm{O}_{2} \text { at STP }
\end{aligned}
$$

60. (D) $\mathrm{PbO}+2 \mathrm{HCl} \longrightarrow \mathrm{PbCl}_{2}+\mathrm{H}_{2} \mathrm{O}$
$x \quad 2 x \quad x$
mole moles mole
Moles of $\mathrm{PbO}=\frac{6.5}{224}$ mole $=0.029$

Moles of $\mathrm{HCl}=\frac{3.2}{36.5}$ mole $=0.087$
Hence limiting reagent is PbO .
And, 0.029 mole of lead (II) chloride will be formed from a reaction between 6.5 g of PbO and 3.2 g of HCl .
61. (C) $3 \mathrm{C}+2 \mathrm{Al}_{2} \mathrm{O}_{3} \longrightarrow 4 \mathrm{Al}+3 \mathrm{CO}_{2}$ (from bauxite)

4 moles of Al is produced by 3 moles of C
1 mole of Al is produced by $\frac{3}{4}$ moles of C
$10^{4}$ moles of Al is produced by $\frac{3}{4} \times 10^{4}$ moles of C
Amount of carbon used $=\frac{3}{4} \times 10^{4} \times 12 \mathrm{~g}$

$$
=90000 \mathrm{~g}=90 \mathrm{~kg}
$$

62. (A) The equation involved in the reaction is

$$
\mathrm{BaCO}_{3} \rightarrow \mathrm{BaO}+\mathrm{CO}_{2} \uparrow
$$

197 g of $\mathrm{BaCO}_{3}$ on decomposition produces 22.4 L of $\mathrm{CO}_{2}$.
$\therefore 9.85 \mathrm{~g}$ of $\mathrm{BaCO}_{3}$ will produce

$$
=\frac{22.4 \times 9.85}{197}=1.12
$$

63. (A) $\quad 4 \mathrm{NH}_{3(g)}+5 \mathrm{O}_{2(g)} \rightarrow 4 \mathrm{NO}_{(g)}+6 \mathrm{H}_{2} \mathrm{O}_{(l)}$

According to equation, 4moles of $\mathrm{NH}_{3}$ reacts with 5 moles of $\mathrm{O}_{2}$.
Hence, 1 mole of $\mathrm{NH}_{3}$ will need $\frac{5}{4}=1.25$ mole of oxygen.
So, $\mathrm{O}_{2}$ is a limiting reagent and it will be consumed completely.
1 mole of $\mathrm{O}_{2}$ will give $=\frac{4}{5}$ moles of $\mathrm{NO}_{(g)}$ and $\frac{6}{5}$ moles of $\mathrm{H}_{2} \mathrm{O}_{(l)}$.
64. (A) For Dilution:

$$
\begin{aligned}
& \mathrm{N}_{1} \mathrm{~V}_{1}=\mathrm{N}_{2} \mathrm{~V}_{2} \\
& \text { Normality }=\text { Molarity } \times \text { Basicity } \\
& \text { For } \mathrm{H}_{2} \mathrm{SO}_{4}, \\
& \text { basicity }=2 \\
& 5 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4}=10 \mathrm{NH}_{2} \mathrm{SO}_{4} \\
& \mathrm{~N}_{1}=10 \mathrm{~N} \\
& \mathrm{~V}_{1}=1 \mathrm{~L} \\
& \mathrm{~N}_{2}=? \\
& \mathrm{~V}_{2}=10 \mathrm{~L} \\
& 10 \times 1=\mathrm{N}_{2} \times 10 \\
& \Rightarrow \quad \mathrm{~N}_{2}=1 \mathrm{~N}
\end{aligned}
$$

## $\triangle$ Caution

Students should keep in mind that for dilution normality term should be used not molarity. If molarity is given in question. convert it to normality.

---

65. (D) The balanced reaction can be written as:

$$
\mathrm{Z}_{2} \mathrm{O}_{3}+3 \mathrm{H}_{2} \rightarrow 2 \mathrm{Z}+3 \mathrm{H}_{2} \mathrm{O}
$$

For 1 mol of $\mathrm{Z}_{2} \mathrm{O}_{3}$, we need 3 moles or 6 g of $\mathrm{H}_{2}$
Thus, 6 mg of $\mathrm{H}_{2}$ is needed for $1 \times 10^{-3} \mathrm{~mol}$ of $\mathrm{Z}_{2} \mathrm{O}_{3}$ ( 0.1596 g given)

Mass of $1 \times 10^{-3} \mathrm{~mol}$ of $\mathrm{Z}_{2} \mathrm{O}_{3}=0.1596 \mathrm{~g}$
So, mass of 1 mol of $\mathrm{Z}_{2} \mathrm{O}_{3}=159.6 \mathrm{~g}$
Let atomic weight of $\mathrm{Z}=x$
Thus,

$$
2 \times x+3 \times 16=159.6
$$

$$
x=\frac{159.6-48}{2}
$$

66. (C) Balanced equation:

$$
\mathrm{C}_{2} \mathrm{H}_{4(g)}+3 \mathrm{O}_{2(g)} \rightarrow 2 \mathrm{CO}_{2(g)}+2 \mathrm{H}_{2} \mathrm{O}_{(l)}
$$

$1 \mathrm{~mol}(28 \mathrm{~g})$ of $\mathrm{C}_{2} \mathrm{H}_{4}$ needs 3 moles or $96 \mathrm{~g} \mathrm{O}_{2}$
Thus, $2.8 \mathrm{~kg} \mathrm{C}_{2} \mathrm{H}_{4}$ will need $9.6 \mathrm{~kg} \mathrm{O}_{2}$
67. (A) Reaction of washing soda $\left(\mathrm{Na}_{2} \mathrm{CO}_{3}\right)$ with $\mathrm{Mg}^{2+}$ ions is:

$$
\begin{aligned}
& \mathrm{Mg}^{2+}+\mathrm{Na}_{2} \mathrm{CO}_{3} \rightarrow \mathrm{MgCO}_{3}+2 \mathrm{Na}^{+} \\
& \text {Milli-equivalents in } 12 \mathrm{mg}=\frac{12 \mathrm{mg}}{12 \mathrm{~g} . \mathrm{eq}} \\
& =1 \text { milli-equivalent } \\
& \text { (g-eq. weight of } \mathrm{Mg}^{2+}=\frac{24}{2}=12 \text { ) }
\end{aligned}
$$

So according to equation, 1 milli-equivalent of $\mathrm{Mg}^{2+}$ will need 1 milli-equivalent of $\mathrm{Na}_{2} \mathrm{CO}_{3}$.

## Related Theory

- Number of g-eq. $=$ Given mass Eq. weight
Eq. weight of ion $=\frac{\text { Molecular weight }}{\text { Charge on ion }}$