Question

A)

A gas mixture with a total pressure of 760 mmHgcontains each of the following gases at the indicated partial pressures: 124 mmHg CO2, 216 mmHg Ar, and 177 mmHg O2. The mixture also contains helium gas.

What is the partial pressure of the helium gas?

Express your answer in millimeters of mercury.

B) A sample of gas has a mass of 0.560 g . Its volume is 116 mL at a temperature of 87 ∘C and a pressure of 769 mmHg .

Find the molar mass of the gas.

Answer 1

A)

total pressure = sum of individual pressure

760 mmHg = p(CO2) + p(Ar) + p (O2) + p(He)

760 mmHg = 124 mmHg + 216 mmHg + 177 mmHg + p(He)

p(He) = 243 mmHg

Answer: 243 mmHg

B)

Given:

P = 769.0 mm Hg

= (769.0/760) atm

= 1.0118 atm

V = 116.0 mL

= (116.0/1000) L

= 0.116 L

T = 87.0 oC

= (87.0+273) K

= 360 K

find number of moles using:

P * V = n*R*T

1.0118 atm * 0.116 L = n * 0.0821 atm.L/mol.K * 360 K

n = 3.971*10^-3 mol

mass(solute)= 0.560 g

use:

number of mol = mass / molar mass

0.004 mol = (0.560 g)/molar mass

molar mass = 141 g/mol

Answer: 141 g/mol