Question

A gas mixture contains nitrogen and helium in 7:4 ratio by weight. The pressure of the mixture is 760 mm. The partial pressure of nitrogen is:

A.0.2 atm
B.0.8 atm
C.0.5 atm
D.0.4 atm

Answer 1

Hint:We know that, Raoult’s law states that for a solution containing volatile solutes, partial vapour pressure of each component of the solution is directly proportional to its mole fraction. Here, we use the relation of ${p_A}\,\alpha \,{x_1}$ (Raoult’s law), where ${p_a}$ represents partial pressure and ${x_1}$ represents mole fraction.

Complete step by step answer:

Let’s first understand the partial pressure. It is the pressure exerted by a gas if it occupies the total volume of a mixture. Partial pressure is dependent on the concentration of gas.

Now from Raoult’s law,

${p_1}\, = p_1^0\,{x_1}$

Where, $p_1^0$ is the vapor pressure of the mixture.

Now, come to the question. A gaseous mixture of two gases namely, nitrogen and helium has vapor pressure 760 mm. Their ratio by mass is 7:4 and we have to calculate the partial pressure of nitrogen.

So, using Raoult’s law,

${p_{{{\rm{N}}_{\rm{2}}}}}\, = p_{{\rm{total}}}^{}\,{x_{{{\rm{N}}_{\rm{2}}}}}$ …… (1)


As the ratio by mass of nitrogen and helium is 7:4, Their masses will be 7x and 4x respectively. Now, we have to calculate the mole fraction of nitrogen. To calculate mole fraction, we need the moles of both nitrogen and helium. The formula to calculate moles is,

${\rm{Number}}\;{\rm{of}}\,{\rm{moles}} = \dfrac{{{\rm{Mass}}}}{{{\rm{Molar}}\,{\rm{mass}}}}$
 …… (2)

First, we calculate moles of nitrogen. Mass of nitrogen is given as 7x and molar mass of nitrogen $\left( {{{\rm{N}}_{\rm{2}}}} \right)$
$ = 14 \times 2 = 28\,\,{\rm{g}}\,{\rm{mo}}{{\rm{l}}^{ - 1}}$.

Using equation (2),

${\rm{Moles}}\;{\rm{of nitrogen}} = \dfrac{{7x}}{{28}} = \dfrac{x}{4}$


Similarly, we have to calculate the moles of helium. Mass of helium is given as 4x and molar mass of helium is $4.00\;{\rm{g}}\,{\rm{mo}}{{\rm{l}}^{ - 1}}$. Using equation (2),

${\rm{Moles}}\;{\rm{of helium}} = \dfrac{{4x}}{4} = x$


Now, expression of mole fraction of hydrogen is,

${x_{{{\rm{N}}_{\rm{2}}}}} = \dfrac{{{\rm{Moles}}\,{\rm{of nitrogen}}}}{{{\rm{Moles}}\,{\rm{of nitrogen}} + {\rm{Moles}}\,{\rm{of helium}}}}$

$ \Rightarrow {x_{{{\rm{N}}_{\rm{2}}}}} = \dfrac{{\dfrac{x}{4}}}{{\dfrac{x}{4} + x}}$

$ \Rightarrow {x_{{{\rm{N}}_{\rm{2}}}}} = \dfrac{{\dfrac{x}{4}}}{{\dfrac{{5x}}{4}}}$

$ \Rightarrow {x_{{N_2}}} = \dfrac{1}{5}$


Now, put the value of ${x_{{N_2}}}$ and total pressure (760 mm) in equation (1).

${p_{{{\rm{N}}_{\rm{2}}}}}\, = p_{{\rm{total}}}^{}\,{x_{{{\rm{N}}_{\rm{2}}}}}$

$ \Rightarrow {p_{{N_2}}} = 760\,{\rm{mm}} \times \dfrac{1}{5} = 152\,{\rm{mm}}$


Now, we have to convert the unit of pressure to atm. The conversion factor is $\dfrac{{1\,{\rm{atm}}}}{{760\,{\rm{mm}}}}$.

So,

$152\,{\rm{mm}} = 152\,{\rm{mm}} \times \dfrac{{1\,{\rm{atm}}}}{{760\,{\rm{mm}}}} = 0.2\,{\rm{atm}}$

Therefore, partial pressure of nitrogen is 0.2 atm.

Hence, the correct answer is option A.


Note:
Remember that, Dalton’s law of partial pressure states that summation of partial pressure of every component of a solution gives the total pressure over the solution phase. For a solution containing only two components 1 and 2, total pressure is ${p_{{\rm{total}}}} = {p_1} + {p_2}$