0.2 g of a primary aliphatic amine, on treatment with nitrous acid gave 61.36 ml of `N_2` gas at STP. The molecular mass of the amine will be:

Options

A. 99 g/mole
B. 87 g/mole
C. 73 g/mole
D. 123 g/mole

Answer

To find the molecular mass of the primary aliphatic amine, we can follow these steps:

### Step 1: Understand the Reaction
The primary aliphatic amine (RNH2) reacts with nitrous acid (HNO2) to produce nitrogen gas (N2) and water. The balanced reaction can be represented as:
\[ RNH_2 + HNO_2 \rightarrow ROH + N_2 + H_2O \]
From this reaction, we can see that 1 mole of amine produces 1 mole of nitrogen gas.

### Step 2: Calculate Moles of Nitrogen Gas
We are given that 61.36 ml of N2 gas is produced at STP (Standard Temperature and Pressure). At STP, 1 mole of gas occupies 22.4 liters (or 22400 ml).

To find the moles of nitrogen gas produced:
\[ \text{Moles of } N_2 = \frac{\text{Volume of } N_2 \text{ (in liters)}}{22.4 \text{ L/mol}} \]
\[ \text{Moles of } N_2 = \frac{61.36 \text{ ml}}{22400 \text{ ml/mol}} \]
\[ \text{Moles of } N_2 = \frac{61.36 \times 10^{-3} \text{ L}}{22.4 \text{ L/mol}} \]
\[ \text{Moles of } N_2 = 2.737 \times 10^{-3} \text{ mol} \]

### Step 3: Relate Moles of Amine to Moles of Nitrogen
Since 1 mole of amine produces 1 mole of nitrogen gas, the moles of amine will also be:
\[ \text{Moles of amine} = 2.737 \times 10^{-3} \text{ mol} \]

### Step 4: Calculate Molar Mass of the Amine
The molar mass (molecular weight) can be calculated using the formula:
\[ \text{Molar Mass} = \frac{\text{Mass of amine (g)}}{\text{Moles of amine (mol)}} \]
Given that the mass of the amine is 0.2 g:
\[ \text{Molar Mass} = \frac{0.2 \text{ g}}{2.737 \times 10^{-3} \text{ mol}} \]
\[ \text{Molar Mass} = 73.0 \text{ g/mol} \]

### Final Answer
The molecular mass of the primary aliphatic amine is **73 g/mol**.

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