Question

0.465 g sample of an unknown compound occupies 245 mL at 298K and 1.22 atm. What is the molar mass of the unknown compound?

Answer 1

The molar mass of the gas is 38.0 g/mol.

One way to solve this problem is to use the Ideal Gas Law.

`PV = nRT`

`n = m/M_"r"`, where `m` is the mass and `M_r` is the relative molar mass. So,

`PV = (m/M_"r")RT`

`M_"r" = (mRT)/(PV) = ("0.465 g" × 0.082 06 cancel("L·atm·K⁻¹")"mol⁻¹" × 298 cancel("K"))/(1.22 cancel("atm") × 0.245 cancel("L")) = "38.0 g/mol"`

The relative molar mass is 38.0 g/mol.

Answer 2

38.4g

Explanation:

`M_r=38.4`

`PV=nRT`

`P=1.22xx1.0xx10^(5)Pa`

`T=298K`

`V=245xx10^(-6)m^(3)`

`R=8.31"J/K/mol"`

`n=(PV)/(RT)=(1.22xx10^(5)xx245xx10^(-6))/(8.31xx298)`

`n=0.0121`

So 0.0121 moles weigh 0.465g

So 1 mole weighs `0.465/0.0121= 38.4"g"`

nb I converted to Pa and used 8.31 for R. It depends which R value you are given.