Hello everyone, welcome back to today's Chemistry session. In this video, we'll be solving the first five questions from the chapter Some Basic Concepts of Chemistry. This chapter forms the foundation of numerical chemistry, so it's very important you not only memorize formulas, but also understand how and when to use them. The questions we solve here will help you build confidence for NEET and JEE level application.

So let's begin right away with Question number 1.

Question 1
"Weight of two moles of the organic compound obtained when sodium ethanoate reacts with sodium hydroxide in presence of calcium oxide is:"

Options are:
A) 30 g
B) 18 g
C) 16 g
D) 32 g

Before solving, let's understand what reaction is happening here.

This is a soda-lime decarboxylation reaction. In this reaction, the sodium salt of a carboxylic acid reacts with sodium hydroxide in the presence of calcium oxide, and the product formed is an alkane with one carbon less.

We use this general reaction pattern:
$\mathrm{RCOONa}+\mathrm{NaOH}(\text { heat, } \mathrm{CaO}) \rightarrow \mathrm{RH}+\mathrm{Na} 2 \mathrm{CO} 3$

In our case, the reactant is sodium ethanoate:

# CH3COONa 

Here, the carbon chain has two carbons. After decarboxylation, one carbon is removed, and hydrogen is added. So the product becomes methane (CH4).

Now the question wants the weight of two moles of methane. Let's calculate it step-by-step.
Molar mass of methane:
$\mathrm{C}=12$

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$\mathrm{H} 4=4 \times 1=4$
Molar mass $=12+4=16 \mathrm{~g} / \mathrm{mol}$

Now use the basic formula:
mass $=$ number of moles $\times$ molar mass

So:
mass $=2 \times 16=32 \mathrm{~g}$

So the correct answer is Option D: 32 grams.

Before we go ahead, one quick reminder: in this type of question, never calculate the mass of the reactant. Always check what the final product is. NEET frequently tests whether students identify the reaction correctly before doing math.

Question 2
"0.24 g of a volatile gas forms 45 mL vapour at NTP. What will be its vapour density?"

Options are:
A) 95.93
B) 99.59
C) 95.39
D) 59.93

Whenever you see a gas problem at NTP, immediately recall:
1 mole of gas at NTP $=22.4 \mathrm{~L}=22400 \mathrm{~mL}$

Now we use a standard gas-related formula to calculate molar mass:
Molar mass $(\mathrm{M})=($ mass $\times 22400) \div$ volume

Now substitute the values:
$M=(0.24 \times 22400) \div 45$
$M=5376 \div 45$
$\mathrm{M} \approx 119.47 \mathrm{~g} / \mathrm{mol}$

Now the question is asking for vapour density, not molar mass.

The formula for vapour density is:

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Vapour density (VD) = molar mass $\div 2$
So:
$V D=119.47 \div 2 \approx 59.73$ which rounds to 59.93
So the correct answer is Option D: 59.93
Shortcut rule to remember:

Molecular mass $=2 \times$ Vapour Density
If you remember this one line, vapour density questions become extremely fast in the exam.

Question 3
"If two substances $A$ and $B$ have equal number of moles, then:"
Options:
A) They contain equal number of atoms
B) They contain equal number of molecules
C) They have equal masses
D) They have equal volumes

To solve this, let's recall the definition of a mole:
1 mole $=6.022 \times 10^{\wedge} 23$ particles

So if two substances have equal moles, they have the same number of molecules, regardless of their chemical formula or molar mass.

Now check the options:

Option A: equal atoms $\rightarrow$ Not necessarily true because different substances may have different number of atoms in one molecule.

Option C: equal mass $\rightarrow$ Wrong because molar masses may differ.
Option D: equal volume $\rightarrow$ Only true if both are gases at the same temperature and pressure.

So the only universally correct statement is:

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Option B: equal number of molecules.

This is a concept-checking question. NEET repeats it often.

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Question 4

"The number of atoms present in 1.12 mL of a diatomic gas at NTP is:"

Options:

A) 3.01 × 10^19
B) 6.02 × 10^19
C) 6.02 × 10^18
D) 3.01 × 10^18

We start by converting volume to moles using the NTP relation again:

Moles = volume ÷ 22400

So:

Moles = 1.12 ÷ 22400
Moles = 0.00005 mol
Or, 5 × 10^-5 mol

Now convert moles to molecules:

Number of molecules = moles × 6.022 × 10^23

So:

Number of molecules = 5 × 10^-5 × 6.022 × 10^23
= 3.011 × 10^19 molecules

Now the gas is diatomic — which means:

1 molecule contains 2 atoms

So:

Number of atoms = molecules × 2
= 3.011 × 10^19 × 2
≈ 6.022 × 10^19 atoms

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So the correct answer is Option B: $6.02 \times 10^{\wedge} 19$ atoms.

One common trap here is students give the number of molecules instead of atoms. Always check whether the gas is monoatomic, diatomic or polyatomic.

Question 5
"How many moles of nitrogen gas are present in $9.03 \times 10^{\wedge} 23$ molecules?"

Options:
A) 0.25 mol
B) 0.50 mol
C) 1.50 mol
D) 2.25 mol

This question directly uses the fundamental mole formula:
Moles = number of molecules $\div$ Avogadro number

So:

Moles $=\left(9.03 \times 10^{\wedge} 23\right) \div\left(6.022 \times 10^{\wedge} 23\right)$

The powers cancel, leaving:
Moles $=9.03 \div 6.022 \approx 1.499 \approx 1.50 \mathrm{~mol}$

So the correct answer is Option C: 1.50 mol .

Final Summary

Let's quickly review all formulas used today:
1 mole $=6.022 \times 10^{\wedge} 23$ particles
Mass $=$ moles $\times$ molar mass
Moles $=$ mass $\div$ molar mass
Moles (at NTP) $=$ volume $\div 22400$
Vapour density $=$ molar mass $\div 2$
Molar mass $=($ mass $\times 22400) \div$ volume

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If these formulas are memorized and you understand when to apply which one, you will solve most mole concept questions in less than a minute.

That completes Questions 1 to 5 with full reasoning. In the next session, we'll continue with the next set and increase the level gradually.