Question

A 0.582 gram sample of a metal, M, reacts completely with sulfuric acid according to:M(s)+H2SO4(aq)--->MSO4(aq)+H2(g).A volume of 269 mL of hydrogen is collected over water; the water level in the collecting vessel is the same as the outside level. Atmospheric pressure is 756.0 Torr and the temperature is 25 °C. Calculate the molar mass of the metal.

Answer 1

M(s)+H2SO4(aq)--->MSO4(aq)+H2(g)

from this balanced equation it is clear that

one mole of metal will produce one mole of H2 gas

The hydrogen that you collected is wet hydrogen it was collected over water

The pressure of the gas collected is the same as the atmospheric pressure 756.0 mmHg

But some of the pressure comes from the water vapour collected along with the hydrogen

At 25°C water vapour pressure is 24.0mmHg

The actual pressure of the H2 collected is 756-24 = 732mmHg = 0.963 atm

now use the equation

PV = nRT

P = pressure of hydrogen gas = 0.963atm

V = volume = 269 mL = 0.269 L

n = moles of H2need to calculate

R = gas constant = 0.082057 L atm/mol-K

T = temp =273 + 25 = 298K

put all these values in the above equation

0.963 x 0.269 = n x 0.0821 x 298

n = 0.259 / 24.4658

n = 0.0106 mole

now we have mass of the metal given 0.582 grams and

moles of the metal = 0.0106 moles

use the formula

moles = mass / molar mass

0.0106 moles = 0.582 g / molar mass

molar mass = 0.582 g / 0.0106 moles

molar mass = 54.906 g / mol