Question

$NaClO$ solution reacts with $H_{2}S{O}_3$, as $NaClO+H_{2}S{O}_3\to NaCl+H_{2}S{O}_4$. A solution of $NaC$l used in the above reaction contained $15g$ of $NaClO$ per litre. The normality of the solution would be

• A. $0.8$
• B. $0.4$
• C. $0.2$
• D. $0.33$

Answer 1

The correct option is B $0.4$

$\text{Normality : It is defined as the number of gram equivalent of solute present in one liter of the solution.}$

$\text{Mathematical expression of normality is:}$

$Normality=\frac{\text{weight of solute}}{\text{equivalent weight of solute}\times \text{Volume of solution(L)}}$

$\text{First we have to calculate the equivalent weight of solute.}$

$\text{The given balanced chemical reaction is,}$

$NaClO+H_{2}S{O}_4\to NaCl+H_{2}S{O}_4$

$\text{From the reaction we conclude that the number of electrons exchanged is, 2 electrons.}$

$\text{Molar mass of solute NaClO = 75 g/mole}$

$\text{Equivalent weigh of olute}=\frac{\text{Molar mass of solute}}{No.ofelectroneexchange}=\frac{75}{2}=37.5g.eq$

$\text{Now we have to calculate the normality of solution.}$

$Normality=\frac{15g}{37.5g.eq\times 1L}$

$\text{Therefore, the correct option is, (a) 0.40}$