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Question
JEE Main
Chemistry

Grignard reagent $\text{RMgBr (P)}$ reacts with water and forms a gas $\text{(Q)}$. One gram of Q occupies $1.4 \text{ dm}^3$ at STP. (P) on reaction with dry ice in dry ether followed by $\text{H}_3\text{O}^+$ forms a compound $\text{(Z)}$. 0.1 mole of $\text{(Z)}$ will weigh _________ g. (Nearest integer)

Identifying Gas Q from Volume and Mass

The Grignard reagent $\text{RMgBr}$ reacts with water to produce an alkane $\text{RH}$ (gas Q) and $\text{Mg(OH)Br}$.

Reaction: $\text{RMgBr} + \text{H}_2\text{O} \rightarrow \text{RH} \uparrow + \text{Mg(OH)Br}$

We are given that 1 gram of gas Q occupies $1.4 \text{ dm}^3$ (or $1.4 \text{ L}$) at Standard Temperature and Pressure (STP).

At STP, the molar volume of any gas is $22.4 \text{ L/mol}$.

Using this information, we can calculate the molar mass of gas Q:

Molar Mass of Q = $ \frac{\text{Mass}}{\text{Volume}} \times \text{Molar Volume at STP} $

Molar Mass of Q = $ \frac{1 \text{ g}}{1.4 \text{ L}} \times 22.4 \text{ L/mol} = 16 \text{ g/mol} $

The alkane with a molar mass of approximately $16 \text{ g/mol}$ is methane ($\text{CH}_4$, Molar Mass = $12 + 4 \times 1 = 16 \text{ g/mol}$).

Therefore, gas Q is methane ($\text{CH}_4$), and the R group in the Grignard reagent $\text{RMgBr}$ is methyl ($\text{CH}_3$).

So, the Grignard reagent (P) is $\text{CH}_3\text{MgBr}$.

Synthesizing Compound Z using Dry Ice

The Grignard reagent $\text{CH}_3\text{MgBr}$ reacts with dry ice ($\text{CO}_2$) followed by acid hydrolysis ($\text{H}_3\text{O}^+$) to form a carboxylic acid.

Step 1: Reaction with $\text{CO}_2$

$ \text{CH}_3\text{MgBr} + \text{CO}_2 \rightarrow \text{CH}_3\text{COOMgBr} $

Step 2: Acid Hydrolysis

$ \text{CH}_3\text{COOMgBr} + \text{H}_3\text{O}^+ \rightarrow \text{CH}_3\text{COOH} + \text{Mg}^{2+} + \text{Br}^- $

The compound Z is acetic acid ($\text{CH}_3\text{COOH}$).

Calculating Weight of Compound Z

We need to find the weight of 0.1 mole of compound Z ($\text{CH}_3\text{COOH}$).

First, calculate the molar mass of $\text{CH}_3\text{COOH}$:

Molar Mass = $ (2 \times \text{Atomic Mass of C}) + (4 \times \text{Atomic Mass of H}) + (2 \times \text{Atomic Mass of O}) $

Molar Mass = $ (2 \times 12.011) + (4 \times 1.008) + (2 \times 15.999) $

Molar Mass = $ 24.022 + 4.032 + 31.998 = 60.052 \text{ g/mol} $

Now, calculate the weight of 0.1 mole of $\text{CH}_3\text{COOH}$:

Weight = Moles $ \times $ Molar Mass

Weight = $ 0.1 \text{ mol} \times 60.052 \text{ g/mol} = 6.0052 \text{ g} $

The question asks for the weight to the nearest integer.

Nearest integer to $6.0052 \text{ g}$ is $6 \text{ g}$.

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