Chapter 10, Problem 68A

Interpretation Introduction

Interpretation: The molar mass of a gas that has density 0.902 g/L should be calculated.

Concept Introduction: Factor that uses one set of units to change it to another set of units is said to be conversion factor.

Answer to Problem 68A

The molar mass of a gas that has density 0.902 g/L is $\text{20}\text{.2 g/mol}$ .

Explanation of Solution

The molar volume of any compound at STP is 22.4 L for 1 mol of the compound.

Now, using the value of density of the given gas and molar volume at STP, the molar mass of the compound can be calculated.

The required conversion factor is:

  $\text{molar mass (g/mol) = }\frac{\text{g}}{\text{L}}\text{×}\frac{\text{22}\text{.4 L}}{\text{1 mol}}$

On substituting the value of density and solving:

  $\text{molar mass (g/mol) = }\frac{\text{0}\text{.902 g}}{\text{L}}\text{×}\frac{\text{22}\text{.4 L}}{\text{1 mol}}\text{ = 20}\text{.2 g/mol}$

Hence, the molar mass of a gas that has density 0.902 g/L is $\text{20}\text{.2 g/mol}$ .