Question

A $1:1$ mixture (by weight) of hydrogen and helium is enclosed in a one-litre flask at $0{}^{\circ }\text{C}$. Assuming ideal behaviour, the partial pressure of helium is found to be $0.42atm$, then the concentration of hydrogen would be:

• A. $0.0375$ M
• B. $0.028$ M
• C. $0.0562$ M
• D. $0.0187$ M

Answer 1

The correct option is A $0.0375$ M

Let the weights of $H_2$ and $He$ in the mixture be $x$ each.
Then the number of moles of $H_2=\frac{x}{2}$
and the number of moles of $He=\frac{x}{4}$

Pressure in the flask $=P=\frac{nRT}{V}$
$P=\frac{(\frac{x}{2}+\frac{x}{4})\times 0.0821\times 273}{1}$
$\Rightarrow P=16.81x$
Partial pressure of $He=0.42\text{atm}$
$\Rightarrow \frac{x/4}{(\frac{x}{2}+\frac{x}{4})}\times P=0.42$
$\Rightarrow \frac{x/4}{3x/4}\times 16.81x=0.42$
$\Rightarrow x=0.075$
$\therefore$ concentration of $H_2=\frac{x/2}{V}=\frac{0.075/2}{1}=0.0375\text{M}$