Question

56 g of nitrogen and 96 g of oxygen are mixed isothermally and at a total pressure of 10 atm. The partial pressures of oxygen and nitrogen (in atm) are _____ respectively

• A. 4, 6
• B. 5, 5
• C. 2, 8
• D. 6, 4

Answer 1

Answer: (D)

6, 4

Raoult's Law states that the partial vapour pressure of a component in a mixture is equal to the vapour pressure of the pure component at that temperature multiplied by its mole fraction in the mixture.

In a mixture of liquids of A and B,

$p_A=x_A\times p$

$p_B=x_B\times p$

moles of nitrogen gas is $n_{N_2}=56/28=2mol$

moles of oxygen gas is $n_{O_2}=96/32=3mol$

Mole fraction of nitrogen is $x_{N_2}=2/(2+3)=0.4$

Mole fraction of oxygen will be $x_{O_2}=1-0.4=0.6$

Partial pressure of oxygen is $p_{O_2}=0.6\times 10=6atm$

Partial pressure of Nitrogen is $p_{N_2}=0.4\times 10=4atm$