# 0.465 g sample of an unknown compound occupies 245 mL at 298K and 1.22 atm. What is the molar mass of the unknown compound?

Mar 20, 2015

The molar mass of the gas is 38.0 g/mol.

One way to solve this problem is to use the Ideal Gas Law.

$P V = n R T$

$n = \frac{m}{M} _ \text{r}$, where $m$ is the mass and ${M}_{r}$ is the relative molar mass. So,

$P V = \left(\frac{m}{M} _ \text{r}\right) R T$

${M}_{\text{r" = (mRT)/(PV) = ("0.465 g" × 0.082 06 cancel("L·atm·K⁻¹")"mol⁻¹" × 298 cancel("K"))/(1.22 cancel("atm") × 0.245 cancel("L")) = "38.0 g/mol}}$

The relative molar mass is 38.0 g/mol.

Mar 20, 2015

38.4g

#### Explanation:

${M}_{r} = 38.4$

$P V = n R T$

$P = 1.22 \times 1.0 \times {10}^{5} P a$

$T = 298 K$

$V = 245 \times {10}^{- 6} {m}^{3}$

$R = 8.31 \text{J/K/mol}$

$n = \frac{P V}{R T} = \frac{1.22 \times {10}^{5} \times 245 \times {10}^{- 6}}{8.31 \times 298}$

$n = 0.0121$

So 0.0121 moles weigh 0.465g

So 1 mole weighs $\frac{0.465}{0.0121} = 38.4 \text{g}$

nb I converted to Pa and used 8.31 for R. It depends which R value you are given.