Question

3.0 g of sample of impure ammonium chloride were boiled with excess of caustic soda solution.Ammonia gas so evolved was passed into 120 ml of N/2 H2SO4. 28 ml of N/2 NaOH were required to neutralize excess of the acid . Calculate the percent purity of the given sample of ammonium chloride

Answer 1

The percentage purity of ammonium chloride is 82.01%

Explanation:

Volume of excess of the acid can be determined as

N1 V1 (Acid) = N2V2 (NaOH)

V1 = N2V2 / N1 = 1/2 × 281 / 2 = 28 ml

Therefore, volume of H2SO4 consumed by NH3 = 120 − 28 = 92 ml

Now, moles of NH3 = moles of H2SO4 consumed

Moles of H2SO4 = Molarity × Volume / 1000

= 12 × 92 / 1000 = 0.046 moles

So, moles of NH3 = 0.046 moles

NH4Cl + NaOH → NH3 + NaCl + H2O

1 mol                      1 mol

1 mol of NH4Cl gives 1 mol of NH3

so, 0.046 moles of NH3 will be given by 0.046 moles of NH4Cl

Therefore, mass of NH4Cl = number of moles×molar mass

                                            = 0.046 × 53.49 = 2.4605 g

This means that 3 g of impure sample contains 2.4605 g of NH4Cl

So, percent purity of given sample = 2.46053 × 100 = 82.01%