Question

A 0.435 g sample of a metal, M, reacts completely with sulfuric acid according to M ( s ) + H 2 SO 4 ( aq ) ⟶ MSO 4 ( aq ) + H 2 ( g ) A volume of 201 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

Answer: The molar mass of metal is 52.4 g/mol

Explanation:

To calculate the moles of hydrogen gas collected, we use the equation given by ideal gas which follows:

$PV=nRT$

where,

P = pressure of the hydrogen gas = Total pressure - vapor pressure of water = (756.0 - 23.8 ) torr = 732.2 torr  

V = Volume of the gas = 201 mL = 0.201 L    (Conversion factor:  1 L = 1000 mL)

T = Temperature of the gas = $25^oC=[25+273]K=298K$

R = Gas constant = $62.364\text{ L.mmHg }mol^{-1}K^{-1}$

n = number of moles of hydrogen gas = ?

Putting values in above equation, we get:

$732.2torr\times 0.210L=n\times 62.364\text{ L. torr }mol^{-1}K^{-1}\times 298K\\\\n=\frac{732.2\times 0.210}{62.364\times 298}=0.0083mol$

The given chemical equation follows:

$M(s)+H_ 2SO_4(aq.)\rightarrow MSO_4(aq.)+H_2(g)$

By Stoichiometry of the reaction:

1 mole of hydrogen gas is formed by 1 mole of metal

So, 0.0083 moles of hydrogen gas will be formed by = $\frac{1}{1}\times 0.0083=0.0083mol$ of metal

To calculate the molar mass of metal from given number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Moles of metal = 0.0083 moles

Given mass of metal = 0.435 g

Putting values in above equation, we get:

$0.0083mol=\frac{0.435g}{\text{Molar mass of metal}}\\\\\text{Molar mass of metal}=\frac{0.435g}{0.0083mol}=52.4g/mol$

Hence, the molar mass of metal is 52.4 g/mol