JEE Chemistry 2026 Question with Solution

Given: Grignard reagent $\mathrm{RMgBr}$ reacts with water to form a gas $Q$. One gram of $Q$ occupies $1.4 \, \text{dm}^3$ at STP. On reaction with dry ice in dry ether followed by $\mathrm{H_3O^+}$, compound $Z$ is formed.

Find: Mass of $0.1$ mole of $Z$.

At STP, $22.4 \, \text{dm}^3$ corresponds to $1$ mole.

So, gas $Q$ is $\mathrm{CH_4}$.

Since hydrolysis of a Grignard reagent gives $\mathrm{RH}$, we get

Therefore, $P$ is $\mathrm{CH_3MgBr}$.

Grignard reagent with $\mathrm{CO_2}$ followed by acidic hydrolysis gives a carboxylic acid. Hence,

Thus, compound $Z$ is $\mathrm{CH_3COOH}$.

Molar mass of $\mathrm{CH_3COOH}$ is

Mass of $0.1$ mole of $Z$ is

Therefore, $0.1$ mole of $Z$ weighs $6 \, \text{g}$.

Given: One gram of gas $Q$ occupies $1.4 \, \text{dm}^3$ at STP, and $\mathrm{RMgBr}$ with dry ice followed by hydrolysis forms compound $Z$.

Find: The required mass of $0.1$ mole of $Z$.

1. Use molar volume at STP:

2. Therefore, molar mass of the gas is

3. A gas of molar mass $16$ is methane, $\mathrm{CH_4}$.

4. Grignard reagent on hydrolysis gives the corresponding alkane $\mathrm{RH}$, so

5. Hence the reagent is $\mathrm{CH_3MgBr}$.

6. Reaction with dry ice gives one-carbon homologated carboxylic acid:

7. The product $Z$ is acetic acid, $\mathrm{CH_3COOH}$, with molar mass $60 \, \text{g mol}^{-1}$.

8. Required mass:

Therefore, the required answer is $6$.