A $10 L$ cylinder contains $N X_{2}$ at $2 7^{\circ} C$ and $10 atm$ . The valve is opened and the gas escapes until the pressure falls to $4 atm$ at the same temperature. Taking $R = 0.0821 L atm K^{- 1} mo l^{- 1}$ , the mass of $N X_{2}$ (molar mass 28) that has escaped is nearest to

Answer: A

At constant

V

and

T

Δ n = \frac{Δ P V}{RT} = \frac{( 10 - 4 ) \times 10}{0.0821 \times 300} = 2.436 mol

. Mass

= 2.436 \times 28 = 68.2 g \approx 68 g