Question

An impure sample of$NaCl$ which weighed $1.2{\text{ }}g$ gave on treatment with excess of $AgN{O_3}$ solution $2.4{\text{ }}g$ of $AgCl$ as precipitate. Calculate the percentage purity of the sample. $\left( {Ag = 108} \right)$

Answer 1

It is well known that when sodium chloride reacts with silver nitrate it results in the formation of silver chloride and sodium nitrate and a white precipitate is formed when silver ions react with chloride ions.

Complete step by step answer:
Let us first write the chemical reaction defining the process of sodium chloride reacting with silver nitrate giving silver chloride and sodium nitrate and a white precipitate which is formed when silver ions react with chloride ions is: $NaCl + AgN{O_3} \to AgCl + NaN{O_3}$ ……(i)
From (i) we can say that 1 mole of $NaCl$ helps in precipitating 1 mole of $AgCl$. In other words $58.5\,g$ of $NaCl$ precipitates in $143.5\,g$ of $AgCl$.
So,$1.2{\text{ }}g$ of $NaCl$ will precipitate $= \dfrac{{143.5}}{{58.5}} \times 1.2$
= 2.94\,g\,$$$AgCl$$ But as you can see in the question it is given that only $$2.4{\text{ }}g$$ of $$AgCl$$ is precipitated so we can get the impurity amount by calculating the difference between precipitate we obtained and given precipitate amount. Thus after subtracting both we get2.94 - 2.4 = 0.54$g which is the impurity amount. Now we can easily calculate the impurity percentage as well as the purity percentage.$
impurity{\text{ }}percentage = \dfrac{{0.54}}{{2.94}} \times 100 \\
{\text{ = 18}}{\text{.4% }}
$$
percentage{\text{ }}purity = {\text{100 - 18}}{\text{.4}} \\
{\text{ = 81}}{\text{.6% }}
$

Alternate method:
Percentage purity tells us about the purity of the substance which can be calculated by simply dividing the mass of the pure chemical by mass of the impure chemical multiplying it by 100 for the percentage.
Now, as the question implies that there is an impure sample of $NaCl$, the first step in finding out the percentage purity is to simply calculate the mass of the pure sample. For this, we should know the molecular masses of the chemicals. So, let us write the molecular mass of $NaCl$ which is $58.5g$ and molecular mass of $AgCl$ which is $143.5\,g$.
Here we can say that $143.5\,g$ of $AgCl$ is obtained from $58.5g$ of pure$NaCl$.
$1\,g{\text{ AgCl = }}\dfrac{{58.5}}{{143.5}} \\\ \Rightarrow 2.4\,g{\text{ AgCl = }}\dfrac{{58.5}}{{143.5}} \times 2.4g \\\ {\text{ = 0}}{\text{.978g}}$
Now, percentage purity can be calculated using the formula:
{\text{percentage purity = }}\dfrac{{{\text{mass of pure sample}}}}{{{\text{mass of impure sample}}}} \times {\text{100}} \\\ {\text{ = }}\dfrac{{0.978}}{{1.2}} \times {\text{100}} \\\ {\text{ = 81}}{\text{.5% }}

Note:
Students should first write the chemical equation for better understanding of the reaction taking place and they can remember the formula of percentage purity and easily calculate it by the first method. And for secondary classes the second method will be preferable for getting good marks.