Question

The partial pressure of hydrogen in a flask containing \[2g\] of ${H_2}$ and $32$ of $S{O_2}$ is:
A $\dfrac{1}{{16}}$ of total pressure
B $\dfrac{1}{2}$ of total pressure
C $\dfrac{2}{3}$ of total pressure
D $\dfrac{1}{8}$ of total pressure

Answer 1

Hint: To find the partial pressure we normally use Dalton’s law. It states that the total pressure exerted by a mixture of gases is equal to the sum of partial pressure of all the gases present in the mixture.

Complete step by step solution:
To find the pressure we generally use ideal gas law which is $P = \dfrac{{nRT}}{V}$where
\[P = \]pressure,$V = $volume $T = $temperature, $n = $number of moles, $R = $universal gas constant.
So here, temperature, volume is constant and universal gas constant already has a constant value
Hence, we can say that $P \propto n$
Let's find the number of moles in a hydrogen molecule as well as sulphur oxide.
Number of mole is $n = \dfrac{w}{{mol.w}}$ where, $w = $weight and $mol.w = $molecular weight
Number of moles of ${H_2}$
$
   = \dfrac{{{w_{{H_2}}}}}{{mol.{w_{{H_2}}}}} \\
   = \dfrac{2}{2} \\
$
Number of moles of $S{O_2}$
$
   = \dfrac{{{w_{S{O_2}}}}}{{mol.{w_{S{O_2}}}}} \\
   = \dfrac{{32}}{{64}} \\
$
Now, let’s find out mole dfraction of hydrogen ${n_{{H_2}}} = \dfrac{{moles{\text{ of }}{{\text{H}}_2}}}{{total{\text{ moles in mixture}}}}$
$\dfrac{{\dfrac{2}{2}}}{{\dfrac{2}{2} + \dfrac{{32}}{{64}}}} = \dfrac{2}{3}$
Partial pressure will be equal to mole dfraction of hydrogen into total pressure
Hence, the partial pressure$ = \dfrac{2}{3}$ total pressure.

Note:
The partial pressure is basically the notional pressure of constituent gas when it is alone occupied by the entire volume of original mixture. Gases basically dissolve, react and diffuse according to partial pressure without any relation with concentration.