Question

$2.26g$ of impure ammonium chloride were boiled with $100mL$ of $NNaOH$ solution till no more ammonia was given off. The excess of $NaOH$ solution left over required $30mL2N{H}_{2}S{O}_4$ for neutralisation. Calculate the percentage purity of the salt.

Answer 1

$N{H}_{4}Cl+NaOH\to NaCl+N{H}_3+H_{2}O$

excess solution of NaOH is neutralised by 30 ml of 2 N $H_{2}S{O}_4$

no . of moles of NaOH neutralised = $\frac{30}{1000}\ast 2$ = 0.06 mole

total moles of NaOH present in 100 ml of N NaOH = $\frac{100}{1000}\ast 1$ = 0.1 mole

so, number of moles of NaOH reacting with ammonium chloride = 0.1 - .06 = 0.04 mole

moles of NaOH = moles of ammonium chloride = 0.04

0.04 moles = $\frac{givenweight}{molecularweight}$

given weight = 2.14 g

percentage purity = $\frac{2.14}{2.26}\ast 100$ = 94.69 %