Question

The density of a gas is $0.259{\text{ g }}{{\text{L}}^{ - 1}}$ at $400{\text{ K}}$ and $190{\text{ torr}}$. Find its molar mass.

Answer 1

Hint: We can solve this question by applying the ideal gas equation. The ideal gas equation gives the relationship between pressure of the gas, volume of the gas and temperature of the gas. Then using the equation of density, substitute the equation of volume in the ideal gas equation and solve for the molar mass.

Formulae Used:
1) $PV = nRT$
2) $d = \dfrac{m}{V}$

Complete step-by-step solution:
We know the ideal gas equation is as follows:
$PV = nRT$ …… (1)
Where $P$ is the pressure of the gas,
$V$ is the volume of the gas,
$n$ is the number of moles of gas,
$R$ is the universal gas constant,
$T$ is the temperature.

We know that the density is the ratio of the molar mass to the volume. Thus,
$d = \dfrac{m}{V}$
Where $d$ is the density of the gas,
$m$ is the molar mass of the gas,
$V$ is the volume of the gas.

Rearrange the equation for the volume of the gas. Thus,
$V = \dfrac{m}{d}$
Thus, equation (1) becomes as follows:
$\Rightarrow P \times \dfrac{m}{d} = nRT$
$\Rightarrow P = \dfrac{d}{m} \times nRT$
Rearrange the equation for the molar mass of the gas. Thus,
$m = \dfrac{d}{P} \times nRT$ …… (2)
Use equation (2) to calculate the molar mass of the gas.
Convert the units of pressure from torr to atm using the relation as follows:
$1{\text{ atm}} = {\text{760 torr}}$
Thus,
$\Rightarrow P = 190{\text{ torr}} \times \dfrac{{1{\text{ atm}}}}{{{\text{760 torr}}}}$
$\Rightarrow P = 0.25{\text{ atm}}$
Thus, the pressure is $0.25{\text{ atm}}$.

Calculate the molar mass of the gas using the equation (2) as follows:
In the equation (2), substitute $0.259{\text{ g }}{{\text{L}}^{ - 1}}$ for the density of the gas, $0.25{\text{ atm}}$ for the pressure of the gas, $0.0821{\text{ L atm mo}}{{\text{l}}^{ - 1}}{\text{ }}{{\text{K}}^{ - 1}}$ or the universal gas constant, $400{\text{ K}}$ for the temperature of the gas. As the number of moles are not given, substitute $1{\text{ mol}}$ for the number of moles of gas and solve for the molar mass of the gas. Thus,
$\Rightarrow m = \dfrac{{0.259{\text{ g }}{{\text{L}}^{ - 1}}}}{{0.25{\text{ atm}}}} \times 1{\text{ mol}} \times 0.0821{\text{ L atm mo}}{{\text{l}}^{ - 1}}{\text{ }}{{\text{K}}^{ - 1}} \times 400{\text{ K}}$
$\Rightarrow m = 34.02{\text{ g mo}}{{\text{l}}^{ - 1}}$

Thus, the molar mass of the gas is $34.02{\text{ g mo}}{{\text{l}}^{ - 1}}$.

Note: The units of pressure should be converted carefully from torr to atm. Remember that $1{\text{ atm}} = {\text{760 torr}}$. Also, use proper units of the universal constant of gas to get the correct answers. The units of universal gas constant should be ${\text{L atm mo}}{{\text{l}}^{ - 1}}{\text{ }}{{\text{K}}^{ - 1}}$.