Question

A 200.0 ml of a gas is collected at 20.0 0C and 733.5mm of Hg. The mass of the sample is 0.934g. What is its Molar mass?

Answer 1

Clapeyron-Mendeleev equation:

$PV=\frac{m}{M}RT$,

where $P$ - pressure, atm; $V$ - volume, l; $m$ - mass of the gas, g; $M$ - molar mass of the gas, g/mol; $R$ - gas constant, $R=0.082 \; atm/(l^3 \cdot K)$; $T$ - temperature, K.

$P[atm]=1.3158 \cdot 10^{−3} \cdot P [mm \; of \; Hg] = 1.3158 \cdot 10^{−3} \cdot 733.5 \approx 0.9651 \; atm$

$T[K]=273.15+T[\degree C]=273.15+20.0=293.15 \; K$

$M=\frac{mRT}{PV}$

$M=\frac{0.934 \cdot 0.082 \cdot 293.15}{0.9651 \cdot 200 \cdot 10^{-3}} \approx 116.32 \; g/mol$

Answer: 116.32 g/mol.