Chemistry 2005 JAMB Past Questions - Question 36: The density of a certain gas is 1.98g/dm-³ at s.tp. What is the molecular mass of the gas?

Correct Answer

A

Explanation

Density = mass/volumeTherefore 1.98 = molecular mass = 1.98 x 22.4 = 44.35 or 44.05
The density of a gas at standard temperature and pressure (STP), which is 0 degrees Celsius (273.15 K) and 1 atmosphere (101.325 kPa), can be used to determine the molar mass of the gas. The ideal gas law, \(PV = nRT\), can be rearranged to solve for the molar mass (\(M\)):

\[ M = \frac{m}{n} \times \frac{V}{RT} \]

where:
- \( M \) is the molar mass of the gas,
- \( m \) is the mass of the gas (given as 1.98 g),
- \( n \) is the number of moles,
- \( V \) is the volume of the gas (in dm³),
- \( R \) is the ideal gas constant (0.0821 dm³⋅mol⁻¹⋅K⁻¹),
- \( T \) is the temperature in Kelvin.

At STP, the volume occupied by one mole of any gas is approximately 22.4 dm³.

Let's calculate the molar mass:

\[ M = \frac{1.98 \, \text{g}}{1 \, \text{mol}} \times \frac{22.4 \, \text{dm}^3}{0.0821 \, \text{dm}^3\text{mol}^{-1}\text{K}^{-1} \times 273.15 \, \text{K}} \]

\[ M \approx \frac{1.98 \, \text{g} \times 22.4 \, \text{dm}^3}{0.0821 \, \text{dm}^3\text{mol}^{-1}\text{K}^{-1} \times 273.15 \, \text{K}} \]

\[ M \approx \frac{44.352 \, \text{g}\cdot\text{dm}^3}{22.414 \, \text{dm}^3\text{mol}^{-1}} \]

\[ M \approx 1.98 \, \text{g/mol} \]

So, the molecular mass of the gas is approximately \(1.98 \, \text{g/mol}\).