A sample of gas at 301 Kelvin and 1.5 atmospheres has a density of 3.90 g/L. What is the molar mass of this gas? Show all of the work used to solve this problem.

PV = nRTP is pressure, V is volume in L, n is number of moles, R is the gas constant,and T is temperature in K

(1.5 atm)(1 L) = (n)(.08206)(301K)

n = .06 moles in one liter

If there are 3.9 grams in .06 moles then

1/.06 x 3.9 = 64.2 grams per mol

Answer Link

Annainn Annainn · Answer 2 · 2019-07-21T09:30:57+02:00

Answer:

Molar mass of the gas is 64.3 g/mol

Explanation:

Given:

Temperature of gas, T = 301 K

Pressure of gas, P = 1.5 atm

Density of gas, D = 3.90 g/L

To determine:

The molar mass, M of the gas

Explanation

Based on the ideal gas equation

[tex]PV = nRT[/tex]

where P = pressure, V = volume ; R = gas constant, T = temperature

n = moles of the gas

[tex]n = \frac{mass(m)}{molar\ mass(M)}[/tex]

substituting for n in the ideal gas equation:

[tex]PV = \frac{m}{M} *RT\\\\M = (\frac{m}{V})*\frac{RT}{P} \\\\density , d = \frac{mass(m)}{Volume(V)} \\\\Therefore:\\\\M = d*\frac{RT}{P} = 3.90 g/L *\frac{.0821Latm/mol-K*301K}{1.5atm} =64.3 g/mol[/tex]