$$\text{RMgI}$$ when treated with ice cold water liberated a gas which occupied $$1.4\,\text{dm}^3/\text{g}$$ at STP. The gas produced is further reacted with iodine in presence of $$\text{HIO}_3$$ to give compound (X). Compound (X) in presence of Na and dry ether produced compound (Y). Molar mass of compound (Y) is $$\underline{\hspace{2cm}}$$ $$\text{g mol}^{-1}$$. (Nearest integer)

Grignard reagents are decomposed by water as

$$\text{RMgI} + \text{H}_2\text{O} \longrightarrow \text{RH (g)} + \text{Mg(OH)I}$$

Hence the only gaseous product is the corresponding alkane $$\text{RH}$$.

Step 1: Determine the molar mass of the liberated gas

Given: the gas occupies $$1.4\,\text{dm}^3$$ per gram at STP.
Density of the gas, $$d = \dfrac{1\,\text{g}}{1.4\,\text{dm}^3} = 0.714\,\text{g dm}^{-3}$$.

Molar mass, $$M = d \times V_\text{m} = 0.714 \times 22.4 \approx 16.0\,\text{g mol}^{-1}$$.

No alkane other than methane has a molar mass close to $$16\,\text{g mol}^{-1}$$, so

$$\text{RH} \equiv \text{CH}_4,\qquad R = \text{CH}_3\!-\!$$

Therefore the original Grignard reagent is $$\text{CH}_3\text{MgI}$$.

Step 2: Reaction with $$\text{I}_2/\text{HIO}_3$$

$$\text{CH}_4 + \text{I}_2 \xrightarrow[\text{HIO}_3]{} \text{CH}_3\text{I} + \text{HI}$$

The oxidant $$\text{HIO}_3$$ regenerates $$\text{I}_2$$ from the HI formed, driving the iodination forward. Hence

Compound $$(X) = \text{CH}_3\text{I}$$ (iodomethane).

Step 3: Wurtz coupling of (X)

In dry ether, sodium couples two molecules of an alkyl halide:

$$2\,\text{CH}_3\text{I} + 2\,\text{Na} \longrightarrow \text{CH}_3{-}\text{CH}_3 + 2\,\text{NaI}$$

Thus compound $$(Y) = \text{C}_2\text{H}_6$$ (ethane).

Step 4: Molar mass of (Y)

$$M_Y = 2(12.0) + 6(1.0) = 30.0\,\text{g mol}^{-1}$$.

The required molar mass of compound (Y) (nearest integer) is 30.