Question

A 280 mL bubble contains 0.283 g of a gas at 0.951 atm and 25.0°C What is the molar mass of this gas?

Answer 1

Answer: 28.3 g/mol

Explanation:

According to the ideal gas equation:'

$PV=nRT$

P= Pressure of the gas = 0.951 atm

V= Volume of the gas = 280 mL = 0.28 L     (1L=1000 ml)

T= Temperature of the gas = 25°C=(25+273)K=298 K   (0°C = 273 K)

R= Value of gas constant = 0.0821 Latm/K mol

$n=\frac{PV}{RT}=\frac{0.951\times 0.28L}{0.0821 \times 298}=0.010moles$

To calculate the moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}$

$0.010=\frac{0.283}{\text {Molar mass}}$

${\text {Molar mass}}=28.3g/mol$

Thus the molar mass of the gas is 28.3 g/mol.