A vessel of volume V $= 20 L$ contains a mixture of hydrogen and helium at a temperature $t = 20^{\circ}$ and pressure p $=$ 2.0 atm. The mass of the mixture is equal to m $=$ 5.0 g. Find the ratio of the mass of hydrogen to that of helium in the given mixture.

0.5

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0.6

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0.7

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0.8

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Solution

The correct option is A 0.5
Let the mixture contain $v_{1}$ & $v_{2}$ moles of $H_{2}$ & $H_{e}$ respectively. If molecular weights of $H_{2}$ & $H_{e}$ are $M_{1}$ & $M_{2}$ , then respective masses in the mixture are equal to
$m_{1} = v_{1} M_{1}$ or $m_{2} = v_{2} M_{2}$
Therefore, for the total mass of the mixture we get,
$m = m_{1} + m_{2}$ or $m = v_{1} M_{1} + v_{2} M_{2}$
Also, if v is the total number of moles of the mixture in the vesses, then we know
$v = v_{1} + v_{2}$
Solving $(1)$ and $(2)$ for $v_{1}$ and $v_{2}$ , we get,
$v_{1} = \frac{(v M_{2} - m)}{M_{2} - M_{1}}, v_{2} = \frac{m - v M_{1}}{M_{2} - M_{1}}$
Therefore, we get $m_{1} = M_{1} \cdot \frac{(v M_{2} - m)}{M_{2} - M_{1}}$ and $m_{2} = M_{2} \frac{(m - v M_{1})}{M_{2} - M_{1}}$
or
$\frac{m_{1}}{m_{2}} = \frac{M_{1}}{M_{2}} \frac{(v M_{2} - m)}{(m - v M_{1})}$
One can also express the above result in terms of the effective molecular weight M of the mixture, defined as,
$M = \frac{m}{v} = m \frac{R T}{p V}$
Thus, $\frac{m_{1}}{m_{2}} = \frac{M_{1}}{M_{2}} \cdot \frac{M_{2} - M}{M - M_{1}} = \frac{1 - M / M_{2}}{M / M_{1} - 1}$
Using the data and table, we get:
$M = 3.0 ℊ$ and, $\frac{m_{1}}{m_{2}} = 0.50$

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