Question

A mixture of dihydrogen and dioxygen at 1 bar pressure contains 20 percent by the weight of dihydrogen. Calculate the partial pressure of dihydrogen.

Answer 1

Mixture of $O_2$ & $H_2$

$P_T=1bar$

$20\%$ by weight of $H_2$

Let the amount of mixture be $100g$

$\Rightarrow m_{H_2}=20g$

& $m_{O_2}=80g$

$n_{H_2}=\frac{m_{H_2}}{2}=\frac{20}{2}=10mol$

$n_{O_2}=\frac{80}{32}=2.5mol$

As, partial pressure is given by

$P_{H_2}=X_H\times P_T$

$X_H=\frac{n_{H_2}}{n_{H_2}+n_{O_2}}=\frac{10}{12.5}=0.8$

$\therefore P_{H_2}=\left(0.8\right)\left(1\right)=0.8bar$

$\therefore$ Partial pressure of $H_2=0.8bar$