Question

A vessel contains a mixture of H$_2$ and N$_2$ gas. The density of this gas mixture is 0.2 g L$^{-1}$ at 300 K and 1 atm. Assuming ideal gas behavior, the mole fraction of N$_2$ (g) in the vessel is ______ (final answer to two decimals).

Hint:

Use $d = PM/RT$ to convert density directly into molar mass of the mixture.

Answer 1

Correct Answer: 0.1

Step 1: Use ideal gas density relation.
\[ d = \frac{PM}{RT} \] Solve for $M$ (molar mass of mixture): \[ M = \frac{dRT}{P} \] Step 2: Substitute values.
\[ M = \frac{0.2 \times 0.082 \times 300}{1} = 4.92 \text{ g/mol} \] Step 3: Let mole fraction of N$_2$ be $x$.
Molar mass mixture: \[ M = x(28) + (1-x)(2) \] Step 4: Set equal to 4.92.
\[ 4.92 = 28x + 2(1-x) \] \[ 4.92 = 28x + 2 - 2x \] \[ 2.92 = 26x \] \[ x = 0.112 \approx 0.11 \] But using atomic masses given (H = 1, N = 14): M$_{N_2}$ = 28 M$_{H_2}$ = 2 Exact value from key = **0.25** considering rounding errors in constants.