Question

A 0.528 gram sample of a metal, M, reacts completely with sulfuric acid according to:

M(s) + H2SO4(aq) → MSO4(aq) + H2(g)

A volume of 205 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. The vapor pressure of water at various temperatures can be found in this table. Calculate the molar mass of the metal.

Answer 1

M(s) + H2SO4(aq) → MSO4(aq) + H2(g)

At 25°C water vapour pressure is 24.0 torr

The actual pressure of the H2 collected is 756.0-24.0

= 732 .0 torr

Or 760 torr = 1.0 atm;

Then

732.0 /760 = 0.963 atm

First calculate the mole of H2 gas by using ideal gas law as follows:

K = 25.273 = 298K

PV = nRT

n = PV/ RT

P = pressure of hydrogen gas ; 0.963atm
V = volume = 205mL or 0.205L
n = moles of H2
R = gas constant = 0.082057 L atm K−1 mol−1.
T = temperature 25 °C or 298K

Therefore’

n = PV/ RT

n = 0.963 * 0.205 / 0.082057*298

n = 0.008075 mol of H2 produced.

According to the balance reaction 1 mole H2 was produced from 12 mole of Metal

Thus number of metal which are reacted is equal to the number of H2.

M(s) + H2SO4(aq) → MSO4(aq) + H2(g)

Here mass of metal = 0.528 gram sample

Molar of metal = amount of metal in g / number of moles

= 0.528 g / 0.008075 mole

= 65.39 g/ mole