Question

The partial pressure of hydrogen in a flask containing 2g of ${{H}_{2}}$ and 32g 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

As we know that if a container filled with a mixture of gases then each gas exerts pressure and the pressure by any gas within the container is termed as its partial pressure. Mathematically, partial pressure of a gas is the product of a mole fraction of the same gas and total pressure of the mixture of gases on the flask or container.
Formula used:
We will use the following formulas:-
$n=\dfrac{w}{M}$
${{x}_{A}}=\dfrac{\text{Moles of component A in the mixture}}{\text{Total moles in a mixture}}$
Partial pressure of component A = ${{x}_{A}}\times \text{Total pressure}$

Complete answer:
Let us first calculate the number of moles of each gas component and then the mole fraction of hydrogen gas followed by the calculation of partial pressure as follows:-
-Calculation of number of moles of ${{H}_{2}}$:-
Given mass of ${{H}_{2}}$in the flask = 2 grams
Molar mass of ${{H}_{2}}$= (1 + 1) g/mol = 2 g/mol
$n=\dfrac{w}{M}$
where,
n = number of moles
w = mass of the element or compound
M = molar mass of the compound.
On substituting the values, we get:-
$\begin{aligned} & \Rightarrow n=\dfrac{w}{M} \\\ & \Rightarrow n=\dfrac{2g}{2g/mol} \\\ & \Rightarrow n=1mole \\\ \end{aligned}$
-Calculation of number of moles of$S{{O}_{2}}$:-
Given mass of $S{{O}_{2}}$in the flask = 32 grams
Molar mass of $S{{O}_{2}}$= (32 + 16 + 16) g/mol = 64 g/mol
On substituting the values, we get:-
$\begin{aligned} & \Rightarrow n=\dfrac{w}{M} \\\ & \Rightarrow n=\dfrac{32g}{64g/mol} \\\ & \Rightarrow n=\dfrac{1}{2}mole \\\ \end{aligned}$
-Calculation of mole fraction of hydrogen in the flask as follows:-
$\begin{aligned} & \Rightarrow {{x}_{{{H}_{2}}}}=\dfrac{\text{Moles of }{{\text{H}}_{2}}\text{ in the mixture}}{\text{Total moles in a mixture}} \\\ & \Rightarrow {{x}_{{{H}_{2}}}}=\dfrac{1}{1+1/2} \\\ & \Rightarrow {{x}_{{{H}_{2}}}}=\dfrac{1}{3/2} \\\ & \Rightarrow {{x}_{{{H}_{2}}}}=\dfrac{2}{3} \\\ \end{aligned}$
-The partial pressure of hydrogen in a flask according to the definition partial pressure is as follows:-
$\begin{aligned} & \Rightarrow {{x}_{{{H}_{2}}}}\times \text{Total pressure} \\\ & \Rightarrow \dfrac{2}{3}\times \text{Total pressure} \\\ \end{aligned}$
-Therefore the partial pressure of hydrogen in a flask containing 2g of ${{H}_{2}}$ and 32g of $S{{O}_{2}}$ is (C) $\dfrac{2}{3}$ of total pressure.

Note:
-Remember that mole fraction is a unit less quantity as both mole units get canceled.
-Also put all the values in the formula along with their units so as to obtain better results with less number of errors and more accuracy which is required in physical chemistry.