Question

2.26 g of impure ammonium chloride was boiled with 100 ml of NaOH solution till no more ammonia was given off. The excess NaOH solution leftover required 30 ml 2N ${{H}_{2}}S{{O}_{4}}$ for neutralization. Calculate the percentage purity of the salt.

Answer 1

Hint: In simple words we can say that the percentage purity of salt is the percentage of each salt element present in the final crystal product. It can be calculated by dividing the total mass of the pure sample and then multiplying it by 100.

Complete step by step answer:
We have studied the concepts of physical chemistry which involve several calculations of the physical quantities like molarity, normality, percent purity, and many more. We shall see the calculation of percent purity in detail.
- We will write the reaction:
\[N{{H}_{4}}Cl\text{+}NaOH\to NaCl+N{{H}_{3}}+{{H}_{2}}O\]
- As we know that excess solution of NaOH is neutralized by 30 ml of 2 N ${{H}_{2}}S{{O}_{4}}$
- So, number of moles of NaOH neutralised will be:
\[\dfrac{{{N}_{1}}\times {{V}_{1}}}{1000}\]
Where, ${{N}_{1}}$ = moles of ${{H}_{2}}S{{O}_{4}}$
${{V}_{1}}$=volume of ${{H}_{2}}S{{O}_{4}}$
We can calculate it as:
$\dfrac{2\times 30}{1000} = 0.06 moles$
Now, we will calculate total moles of NaOH present in 100 ml of N NaOH=
$\dfrac{100}{1000}\times 1 = 0.1 moles$
-So, we can write the number of moles of NaOH reacting with ammonium chloride=
moles of NaOH – moles of ammonium chloride
= 0.1 - 0.06
=0.04 moles
- Percentage purity of the salt = $\dfrac{given\text{ }weight}{molecular\text{ }weight}$
Given weight is = 2.14g
Given molecular weight = 2.26g
Percentage purity = $\dfrac{2.14}{2.26}\times 100=94.96$%
- Hence, we can say that the percentage purity of the salt is 94.69%
So, the correct answer is “Option D”.

Note: It is found that percentage purity is important. As we know that if the product is more processed by crystallization or distillation, the process will be costlier, but we get the purer product. So, it is important that before the sale, the product is analyzed as to its percentage purity.