Question

A $821{\text{ mL}}$ ${{\text{N}}_2}\left( {\text{g}} \right)$ was collected over liquid water at $300{\text{ K}}$ and $1{\text{ atm}}$. If vapour pressure of ${{\text{H}}_{\text{2}}}{\text{O}}$ is $30{\text{ torr}}$ then moles of ${{\text{N}}_2}\left( {\text{g}} \right)$ in moist gas mixture is:
A) 0.39
B) 0.032
C) 0.96
D) 0.0013

Answer 1

To solve this we first have to calculate the vapour pressure of the nitrogen gas. The vapour pressure of ${{\text{N}}_2}$ gas can be calculated by subtracting the vapour pressure of liquid water from the total pressure. Then using the ideal gas equation we can calculate the number of moles of nitrogen.

Formulae Used: $PV = nRT$
Complete step-by-step answer: We are given that the volume of gas is $821{\text{ mL}}$. Thus, $V = 821{\text{ mL}}$ and the pressure of the gas is $1{\text{ atm}}$. Thus, $P = 1{\text{ atm}}$.
Vapour pressure of liquid water is $30{\text{ torr}}$. Thus, ${P_{{{\text{H}}_2}{\text{O}}}} = 30{\text{ torr}}$.
We know that $1{\text{ atm}} = 760{\text{ torr}}$. Thus, ${P_{{{\text{H}}_2}{\text{O}}}} = 30{\text{/760 atm}} = 0.0395{\text{ atm}}$
The temperature of the gas is $300{\text{ K}}$. Thus, $T = 300{\text{ K}}$.
Calculate the pressure of ${{\text{N}}_2}$ gas as follows:
The vapour pressure of ${{\text{N}}_2}$ gas can be calculated by subtracting the vapour pressure of liquid water from the total pressure. Thus,
${P_{{{\text{N}}_2}}} = P - {P_{{{\text{H}}_2}{\text{O}}}}$
Substitute $760{\text{ torr}}$ for the total pressure, $30{\text{ torr}}$ for the vapour pressure of liquid water. Thus,
${P_{{{\text{N}}_2}}} = \left( {1 - 0.0395} \right){\text{ atm}}$
${P_{{{\text{N}}_2}}} = 0.9605{\text{ atm}}$
Thus, the vapour pressure of ${{\text{N}}_2}$ gas is $0.9605{\text{ atm}}$.
Calculate the number of moles of ${{\text{N}}_2}$ gas using the ideal gas equation.
We know that the expression for the ideal gas law is as follows:
$PV = nRT$
Where, $P$ is the pressure of the gas,
$V$ is the volume of the gas,
$n$ is the number of moles of gas,
$R$ is the universal gas constant,
$T$ is the temperature of the gas.
Rearrange the equation for the number of moles,
$n = \dfrac{{PV}}{{RT}}$
Substitute $0.9605{\text{ atm}}$ for the pressure of ${{\text{N}}_2}$ gas, $821{\text{ mL}} = 0.821{\text{ L }}$ for the volume of the gas, $0.0821{\text{ L atm mo}}{{\text{l}}^{ - 1}}{\text{ }}{{\text{K}}^{ - 1}}$ for the universal gas constant, $300{\text{ K}}$ for temperature. Thus,
$n = \dfrac{{0.9605{\text{ atm}} \times 0.821{\text{ L}}}}{{0.0821{\text{ L atm mo}}{{\text{l}}^{ - 1}}{\text{ }}{{\text{K}}^{ - 1}} \times 300{\text{ K}}}}$
$n = 0.032{\text{ mol}}$
Thus, the moles of ${{\text{N}}_2}\left( {\text{g}} \right)$ in a moist gas mixture is 0.032.

Thus, the correct option is (B) 0.032.

Note: The vapour pressure is the tendency of a material to transform into gaseous state or vapour state. The vapour pressure increases as the temperature increases. The pressure exerted by vapours which are in a thermodynamic equilibrium with the condensed phase is known as the vapour pressure.