Question

A 821 mL ${N_2}\left( g \right)$ was collected over liquid water at 300 K and 1 atm. If vapour pressure of ${H_2}O$ is 30 torr then moles of ${N_2}\left( g \right)$ in moist gas mixture is:
(A) $0.39$
(B) $0.032$
(C) $0.96$
(D) $0.0013$

Answer 1

Ideal gas law is an equation of state of ideal gas. Ideal gas is a hypothetical gas. Ideal gas is a theoretical gas composed of many randomly moving point particles. It can be represented by the formula given.
PV = nRT
where P is the pressure, V is the volume, T is the temperature, n is the moles of the gas or the amount of substance and R is the ideal gas constant.

Complete Step by step solution:
The total pressure of the gas mixture is equal to 1 atm which is equal to 760 torr. The volume of the gas collected is 821 mL or $0.821$ L. The vapour pressure of ${H_2}O$ is 30 torr. The pressure of ${N_2}$ gas is equal to the pressure of the gas mixture subtracted by the vapour pressure of water that is 760 – 30 torr which is equal to 730 torr.
730 torr = $\dfrac{{730}}{{760}}$ atm
The value of R is constant for all the gases. And the temperature given in the question is 300 K.
Now, PV = nRT
Hence n = $\dfrac{{{\text{PV}}}}{{{\text{RT}}}} = \dfrac{{\dfrac{{730}}{{760}} \times 0.821}}{{0.0821 \times 300}} = \dfrac{{73}}{{2280}} = 0.032$ moles
Therefore, the total moles of the nitrogen gas in moist gas mixture will be $0.032$ moles.
Hence, the correct answer is option B.

Note:
The value of ideal gas constant depends on the units that are considered in the reaction. Real gas obeys van der waals equation. Real gas has attractive and repulsive forces while ideal gas does not possess any force.