Question

A mixture of nitrogen and xenon gases, at a total pressure of 836 mm Hg, contains 2.80 grams of nitrogen and 24.9 grams of xenon. What is the partial pressure of each gas in the mixture?

Answer 1

Answer: Partial pressure of nitrogen and xenon are 288mmHg and 548 mmHg respectively.

Explanation:

The partial pressure of a gas is given by Raoult's law, which is:

$p_A=p_T\times \chi_A$

where,

$p_A$ = partial pressure of substance A

$p_T$ = total pressure

$\chi_A$ = mole fraction of substance A

We are given:

$m_{N_2}=2.80g$

$m_{Xe}=24.9g$

Mole fraction of a substance is given by:

$\chi_A=\frac{n_A}{n_A+n_B}$

And,

$n_A=\frac{m_A}{M_A}$

Mole fraction of nitrogen is given as:

$\chi_{N_2}=\frac{\frac{m_{N_2}}{M_{N_2}}}{(\frac{m_{N_2}}{M_{N_2}}+\frac{m_{Xe}}{M_{Xe}})}$

Molar mass of $N_2$ = 28 g/mol

Molar mass of $Xe$ =  g/mol

Putting values in above equation, we get:

$\chi_{N_2}=\frac{\frac{2.80}{28}}{\frac{2.80}{28}+\frac{24.9}{131}}$

$\chi_{N_2}=\frac{0.100}{0.100+0.190}=0.345$

To calculate the mole fraction of xenon, we use the equation:

$\chi_{Xe}+\chi_{N_2}=1\\\\\chi_{Xe}=1-0.345=0.655$

$p_{N_2}=836mmHg\times 0.345=288mmHg$

$p_{Xe}=836mmHg\times 0.655=548mmHg$

Thus partial pressure of nitrogen and xenon are 288mmHg and 548 mmHg respectively.