## Question

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

## Answers

Vapor pressure of water at 20

^{o}C = 17.5 mmHgPartial pressure of H

_{2}gas = (Atmospheric pressure) - (Vapor pressure of water at 20^{o}C)Partial pressure of H

_{2}gas = (760 mmHg) - (17.5 mmHg)

Partial pressure of H_{2}gas = 742.5 mmHgPartial pressure of H

_{2}gas = 742.5 mmHg * (1 atm / 760 mmHg)

Partial pressure of H_{2}gas = 0.977 atm(a) Using ideal gas law PV = nRT

n = PV / RT

moles H

_{2}produced = [(pressure H_{2}) * (volume H_{2})] / [(R) * (Temperature H_{2})]where volume H

_{2}= 8 mL = 8 x 10^{-3}LR = gas constant = 0.0821 L-atm/mol-K

Temperature H

_{2}= 20^{o}C = 293 KSubstituting the values,

moles H

_{2}produced = [(0.977 atm) * (8 x 10^{-3}L)] / [(0.0821 L-atm/mol-K) * (293 K)]

moles H_{2}produced = 3.25 x 10^{-4}mol(b) moles X = (moles H

_{2}produced) * (1 mole X / 1 mole H_{2})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