Question

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.

Answer 1

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 2

Molar mass of the gas is 64.3 g/mol

Step-by-step 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

`PV = nRT`

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

n = moles of the gas

`n = (mass(m))/(molar\ mass(M))`

substituting for n in the ideal gas equation:

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