BITSAT 2015 Paper

Let there are $n_1$ moles of hydrogen and $n_2$ moles of helium in the given mixture. As Pv = nRT Then the pressure of the mixture P = $\frac{n_1RT}{V}+\frac{n_2RT}{V}$ = $(n_1+n_2)\frac{RT}{V}$ ⇒ 2 × 101.3 × $10^3$ = $(n_1+n_2)$ × $\frac{8.3\times300}{20\times10^{-3}}$ or $(n_1+n_2)$ = $\frac{2\times101.3\times10^3\times20\times10^{-3}}{(8.3)(300)}$ or $n_1+n_2$ = 1.62 ... (1) The mass of the mixture is (in grams) $n_1$ × 2 + $n_2$ × 4 = 5 ⇒ $(n_1+2n_2)$ = 2.5 ... (2) Solving the eqns. (1) and (2), we get $n_1$ = 0.74 and $n_2$ = 0.88 Hence $\frac{m_H}{m_{He}}$ = $\frac{0.74\times2}{0.88\times2}$ = $\frac{1.48}{3.52}$ = $\frac{2}{5}$