Question 11.8: A sample of gas has a mass of 0.311 g. Its volume is 0.225 L......

GIVEN:

m = 0.311 g

V = 0.225 L

t = 55 °C

P = 886 mm Hg
FIND: Molar mass (g/mol)

RELATIONSHIPS USED

PV = nRT (Ideal gas law, presented in this section)

Molar mass = \frac{\operatorname{Mass}\left(m\right)}{\operatorname{Moles}\left(n\right)} (Definition of molar mass from Section 6.3)

PV = nRT

T = 55 °C + 273 = 328 K

= 9.7\underline{4}06 × 10^{-3}  mol

Molar mass = \frac{\operatorname{Mass}\left(m\right)}{\operatorname{Moles}\left(n\right)} \\ =\;\frac{0.311  g}{9.7\underline{4}06}\times10^{-3}\,\mathrm{mol}

= 31.9 g/mol