Chapter 13, Problem 70A

Interpretation Introduction

Interpretation:

The molar mass of a compound is to be calculated when pressure, temperature and vapor density are given.

Concept introduction:

The equation which helps to calculate density of a gas at given temperature and pressure is given by,

$\begin{array}{l}\text{D=}\frac{\text{PM}}{\text{RT}}\\ \text{where}\\ \text{D}\text{is}\text{the}\text{density}\text{of}\text{gas}\\ \text{M}\text{is}\text{the}\text{molecular}\text{mass}\text{of}\text{gas}\end{array}$

$\mathrm{P}$ is pressure of the gas

$\mathrm{R}$ is the ideal gas constant

$\mathrm{T}$ is the temperature of the gas

Answer to Problem 70A

The molar mass of geraniol is $\text{150}\text{.03}\text{g/mol}$

Explanation of Solution

The formula to calculate density is given by:

$\begin{array}{l}\text{D=}\frac{\text{PM}}{\text{RT}}\\ \therefore \text{M=}\frac{\text{DRT}}{\text{P}}\end{array}$

According to the statement:

$\begin{array}{l}\text{P=0}\text{.140}\text{atm}\\ \text{D=0}\text{.480}\text{g/L}\\ \text{R=0}\text{.0821}{\text{LatmK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\\ {\text{T}}_{}\text{=260}\text{°C}\\ \text{=260+273}\\ \text{=533}\text{K}\end{array}$

Now substituting the values into formula,

$\begin{array}{c}\text{M=}\frac{\text{DRT}}{\text{P}}\\ =\frac{0.480\times 0.0821\times 533}{0.140}\\ \text{=150}\text{.03}\text{g/mol}\end{array}$

Therefore density of the gas is $\text{4}\text{.55}\text{g/L}$

Conclusion

The molar mass of geraniol is $\text{150}\text{.03}\text{g/mol}$.