Question

A 1.00 L sample of a gas at 25.0◦C and 1.00 atm contains 50.0 % helium and 50.0 % neon by mass. What is the partial pressure of the neon. Express your answer in atmospheres.

Answer 1

Hope u will be understand

Answer 2

Answer:

The partial pressure of the neon gas is 0.16 atm.

Explanation:

50.0 % helium and 50.0 % neon by mass.

Total pressure of the gas = P = 1 atm

Let the mass of the gases be M

Moles of helium = $\frac{0.5M}{4 g/mol}$

Moles of neon = $\frac{0.5M}{20 g/mol}$

Using ideal gas equation;

$PV=nRT$

$n=\frac{PV}{RT}=\frac{1.00 L\times 1.00 atm}{0.0820 L atm/mol K\times 298 K}=0.0409 mol$

n = total number of moles of gases

$0.0409 mol=\frac{0.5M}{4 g/mol}+\frac{0.5M}{20 g/mol}$

M = 0.2728 g

Mass of helium gas = $\frac{0.2728g}{2}=0.1364 g$

Moles of helium gas =$n_1=\frac{0.1364 g}{4 g/mol}=0.0341 mole$

Mass of neon gas = $\frac{0.2728g}{2}=0.1364 g$

Moles of neon gas = $n_2=\frac{0.1364 g}{20 g/mol}=0.00682mole$

Partial pressure of the neon gas:

$P\times \chi_2=P\times \frac{n_2}{n_1+n_2}=1 atm\times \frac{0.00682}{0.00682+0.0341}=0.16 atm$

The partial pressure of the neon gas is 0.16 atm.