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# The mixture of $NaI$ and $NaCl$ in reaction with ${H_2}S{O_4}$ gave $N{a_2}S{O_4}$ equal to weight of original mixture taken. The percentage of $NaI$ in the mixture is:A. $82.38$B. $26.38$C. $62.38$D. $28.38$

Last updated date: 20th Jun 2024
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Hint: Percentage of substance in a solution is calculated as the mass of that substance divided by the total mass of the solution multiplied by hundred. And the number of moles of the substance is calculated as the mass of the substance divided by the molar mass of the substance.

Molar mass of substance: It is defined as the mass of one mole of a substance.
Number of mole: It is defined as the ratio of mass of substance to the molar mass of the substance.
In the question,
Let the mass of sodium iodide be $xgm$and mass of sodium chloride be $ygm$.
And reaction of sodium iodide and sodium chloride with sulphuric acid to form sodium sulphate is as:
$2NaI + {H_2}S{O_4} \to N{a_2}S{O_4} + 2HI$
$2NaCl + {H_2}S{O_4} \to N{a_2}S{O_4} + 2HCl$
We know that molar mass of sodium iodide is $127 + 23 = 150gm$
Molar mass of sodium chloride is $35.5 + 23 = 58.5gm$
Molar mass of sodium sulphate is $23 \times 2 + 32 + 16 \times 4 = 142gm$
From the reaction it is clear that two moles of sodium iodide will form one mole of sodium sulphate and similarly two moles of sodium chloride will form one mole of sodium sulphate.
Now the total mass of sodium sulphate will be weight of the original mixture i.e. $(x + y)gm$.
Moles of sodium iodide is $\dfrac{x}{{150}}moles$and moles of sodium chloride is $\dfrac{y}{{58.5}}moles$.
Total moles of sodium sulphate will be as: $(\dfrac{x}{{150}} + \dfrac{y}{{58.5}}) \times \dfrac{1}{2}$.
If from moles we calculate mass then it will be as $(\dfrac{x}{{150}} + \dfrac{y}{{58.5}}) \times \dfrac{1}{2} \times 142$. And also in question the total mass is given as $x + y$.
So, equating these two we will get equation as:
$(\dfrac{x}{{150}} + \dfrac{y}{{58.5}}) \times \dfrac{1}{2} \times 142 = x + y \\ \dfrac{{71x}}{{150}} + \dfrac{{71y}}{{58.5}} = x + y \\ 0.52x = 0.21y \\$
Now, the percentage of sodium iodide is as: mass of sodium iodide divides mass of both sodium iodide and sodium chloride multiply by $100$.
$\%$ of $NaI$$= \dfrac{x}{{x + y}} \times 100. Already we have calculated the relation between x and y i.e. 0.52x = 0.21y. So, \% of NaI$$ = \dfrac{x}{{x + \dfrac{{0.52x}}{{0.21}}}} \times 100$$= 28.4$. The percentage of sodium iodide is $28.4\%$ so the percentage of sodium chloride in the mixture will be $100 - 28.4 = 71.6\%$.

So, the correct answer is Option D.

Note:
There is no need to calculate the values of $x$and $y$separately when you have to find the percentage because in percentage we have to find the ratio. Percentage of $x$ is as $\dfrac{x}{{x + y}} \times 100$and percentage of $y$ is as $\dfrac{y}{{x + y}} \times 100$. So one relation is sufficient to calculate the percentage.