A 0.595 g sample of a metal, M, reacts completely with sulfuric acid according to the reaction M (s) + H 2 SO 4 (aq) ⟶ MSO 4 (aq) + H 2 (g) A volume of 231 mL of hydrogen gas is collected over water; the water level in the collecting vessel is the same as the outside level. Atmospheric pressure is 1.0079 bar and the temperature is 25 °C. The vapor pressure of water at 25 °C is 0.03167 bar. Calculate the molar mass of the metal.

molar mass M (s) = 65.326 g/mol

Explanation:

M (s) + H2SO4 (aq) → MSO4 (aq) + H2 (g)

∴ VH2 (g) = 231 mL = 0.231 L

∴ P atm = 1.0079 bar

∴ PvH2O (25°C) = 0.03167 bar

Graham's law:

⇒ PH2 (g) = P atm - PvH2O (25°C)

⇒ PH2 (g) = 1.0079 bar - 0.03167 bar = 0.97623 bar = 0.9635 atm

∴ nH2 (g) = PV/RT

⇒ nH2 (g) = ((0.9635 atm) (0.231 L)) / ((0.082 atmL/Kmol) (298 K))

⇒ nH2 (g) = 9.1082 E-3 mol

⇒ n M (s) = (9.1082 E-3 mol H2 (g)) (mol M (s) / mol H2 (g))

⇒ n M (s) = 9.1082 E-3 mol

∴ molar mass M (s) [=] g/mol

⇒ molar mass M (s) = (0.595 g) / (9.1082 E-3 mol)

⇒ molar mass M (s) = 65.326 g/mol