Question

A mixture (by weight) of hydrogen and helium is enclosed in a two-litre flask at ${27}^{o}C$. Assuming ideal kept behaviour, the partial pressure of helium is found to be 0.2atm. Then the concentration of hydrogen would be:

• A. 0.045
• B. 8.26
• C. 0.0162
• D. 1.62

Answer 1

The correct option is C 0.0162

$\text{Given,}{H}_2\text{and He gas mixture is enclosed in a flask.}$

$\begin{array}{c}V=2L\\ T=300K\end{array}$

$\text{It is given, Partial pressure of He is}0.2atm\text{.}$

$\text{so,}{n}_{\text{He}}=\frac{P_{\text{He}}V}{RT}=\frac{0.2\times 2}{0.0821\times 300}\text{moles}$

$=0.01624\text{moles.}$

$\text{Since, mixture constitutes}1:1H_2\text{- He ratio by weight.}$

$\text{Thus, weight of}{H}_2=0.01624\times 4gm=0.06496gm$

$\therefore \text{moles of}{H}_2=0.03248mol$

$\text{so, concentration of}{H}_2=\frac{\text{moles}}{\text{volume}}=\frac{0.03248}{2}=0.01624M$