RMgI when treated with ice cold water liberated a gas which...

The reaction of Grignard reagent with water yields an alkane:

$RMgI + H_2O \rightarrow RH + Mg(OH)I$

Given that $1 \text{ g}$ of the gas $RH$ occupies $1.4 \text{ dm}^3$ at STP, we can find its molar mass. Using the standard molar volume at STP ($22.4 \text{ dm}^3$):

$\text{Molar mass of } RH = \dfrac{22.4 \text{ dm}^3 \text{ mol}^{-1}}{1.4 \text{ dm}^3 \text{ g}^{-1}} = 16 \text{ g mol}^{-1}$

The alkane with a molar mass of $16 \text{ g mol}^{-1}$ is methane ($CH_4$).

Methane reacts with iodine in the presence of an oxidizing agent like $HIO_3$ (which prevents the backward reaction by consuming the formed $HI$) to give methyl iodide:

$CH_4 + I_2 \xrightarrow{HIO_3} CH_3I + HI$

Thus, compound (X) is methyl iodide ($CH_3I$).

Compound (X) undergoes the Wurtz reaction when treated with sodium and dry ether to form a higher alkane:

$2CH_3I + 2Na \xrightarrow{\text{dry ether}} CH_3-CH_3 + 2NaI$

Compound (Y) is ethane ($C_2H_6$).

The molar mass of ethane ($C_2H_6$) is:

$2 \times 12 + 6 \times 1 = 30 \text{ g mol}^{-1}$

Answer: $30$