X g of ethylamine is subjected to reaction with NaNO₂/HCl followed by water; evolved dinitrogen gas which occupied 2.24 L volume at STP. X is ______ × 10^–1 g.

The correct answer is 45.

1. Understanding the Reaction:
When a primary aliphatic amine like ethylamine (_C2^H5^NH2) reacts with nitrous acid (HNO₂, prepared in situ using NaNO₂ and HCl), it forms an unstable aliphatic diazonium salt. This salt immediately decomposes in the presence of water to yield ethanol, nitrogen gas (N₂), and hydrochloric acid.

The balanced chemical equation for the reaction is:
$C_2{H}_5{\mathrm{NH}}_2+{\mathrm{HNO}}_2\to C_2{H}_5\mathrm{OH}+N_2+H_{2}O$
From the equation, 1 mole of ethylamine (C₂H₅NH₂) reacts to produce 1 mole of dinitrogen gas (N₂).

2. Calculating the Moles of Nitrogen Gas Evolved:
At Standard Temperature and Pressure (STP), 1 mole of any ideal gas occupies a volume of 22.4 L.
Given that the volume of N₂ evolved is 2.24 L, we can find the number of moles of N₂ gas:
$\text{Moles of}{N}_2=\frac{\text{Volume of gas at STP}}{\text{Molar volume at STP}}$
$\text{Moles of}{N}_2=\frac{2.24\text{L}}{22.4\text{L/mol}}=0.1\text{mol}$

3. Calculating the Mass of Ethylamine (X):
Since the stoichiometric ratio between ethylamine and nitrogen gas is 1:1, the number of moles of ethylamine reacted is also 0.1 mol.

Now, let us calculate the molar mass of ethylamine (C₂H₅NH₂):
$\text{Molar mass of}{C}_2{H}_5{\mathrm{NH}}_2=(2\times 12)+(5\times 1)+14+(2\times 1)$
$\text{Molar mass of}{C}_2{H}_5{\mathrm{NH}}_2=24+5+14+2=45\text{g/mol}$

The mass (X) of ethylamine is calculated as:
$X=\text{Moles}\times \text{Molar mass}$
$X=0.1\text{mol}\times 45\text{g/mol}=4.5\text{g}$

4. Expressing the Answer in the Required Format:
We need to write the mass in the form of X = ______ × 10^-1 g.
$4.5\text{g}=45\times {10}^{-1}\text{g}$
Thus, the value of the integer is 45.