Question

An unknown gas at 21.1 °C and 1.00 atm has a molar mass of 16.04 g/mol.

Assuming ideal behavior, what is the density of the gas?

Answer 1

Answer 2

Given:

• Temperature ($T$) = 21.1 °C = 294.25 K (converted: $21.1+273.15$)

• Pressure ($P$) = 1.00 atm

• Molar mass ($M$) = 16.04 g/mol

• Ideal gas constant ($R$) = 0.0821 L·atm/(mol·K)

• 
  

Formula (Ideal Gas Law for Density):

$$\rho =\frac{PM}{RT}$$

Calculation:

$$\rho =$$$$\rho =\frac{16.04}{24.17}\approx 0.6636\text{g/L}$$

Answer:

The density of the gas is 0.664 g/L (rounded to 3 decimal places).