Experiment 12: Generating Hydrogen Gas Part B: Molar mass of unknown metal Unknown #: Mass of...

Vapor pressure of water at 20 ^oC = 17.5 mmHg

Partial pressure of H₂ gas = (Atmospheric pressure) - (Vapor pressure of water at 20 ^oC)

Partial pressure of H₂ gas = (760 mmHg) - (17.5 mmHg)

Partial pressure of H₂ gas = 742.5 mmHg

Partial pressure of H₂ gas = 742.5 mmHg * (1 atm / 760 mmHg)

Partial pressure of H₂ gas = 0.977 atm

(a) Using ideal gas law PV = nRT

n = PV / RT

moles H₂ produced = [(pressure H₂) * (volume H₂)] / [(R) * (Temperature H₂)]

where volume H₂ = 8 mL = 8 x 10^-3 L

R = gas constant = 0.0821 L-atm/mol-K

Temperature H₂ = 20 ^oC = 293 K

Substituting the values,

moles H₂ produced = [(0.977 atm) * (8 x 10^-3 L)] / [(0.0821 L-atm/mol-K) * (293 K)]

moles H₂ produced = 3.25 x 10^-4 mol

(b) moles X = (moles H₂ produced) * (1 mole X / 1 mole H₂)

moles X = (3.25 x 10^-4 mol) * (1 / 1)

moles X = 3.25 x 10^-4 mol

(c) molar mass of metal = (mass of metal) / (moles of metal)

molar mass of metal = (0.29 g) / (3.25 x 10^-4 mol)

molar mass of metal = 892.5 g/mol