Question

135 mL of a gas is collected over water at 25ºC and 0.993 bar. If the gas weighs 0.160 g and the aqueous tension at 25ºC is 0.0317, calculate the molar mass of the gas.

Answer 1

Given V = 135 mL = 0.135 L

T = 25 +273 = 298 K

P_Total = 0.993 bar

P_H2O = 0.0317 bar

M =?

m = 0.160 g

P_gas = P_Total - P_H2O

= 0.993 - 0.0317 = 0.9613 bar

According to ideal gas equation,

PV = nRT

PV = \(\frac{m}{M}\)RT

\(\therefore\) M = \(\frac{mRT}{PV}\)

= \(\frac{0.160\,g\times0.083\,bar\,mol^{-1}K^{-1}\times298\,K}{0.9613\,bar\times0.135\,L}\)

= 30.49 g mol^-1