Question

2. NaClO solution reacts with H2SO3 as,
NaClO + H2SO3 ---->
NaCl + H S04
A solution of NaClO used in the above reaction contained
15g of NaClO per litre. The normality of the solution would
be :
(a) 0.40 (b) 0.20 (c) 0.60 (d) 0.80
1.
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Answer 1

Answer : The correct option is, (a) 0.40

Explanation :

Normality : It is defined as the number of gram equivalent of solute present in one liter of the solution.

Mathematical expression of normality is:

$\text{Normality}=\frac{\text{Gram equivalent of solute}}{\text{Volume of solution in liter}}$

or,

$\text{Normality}=\frac{\text{Weight of solute}}{\text{Equivalent weight of solute}\times \text{Volume of solution in liter}}$

First we have to calculate the equivalent weight of solute.

The given balanced chemical reaction is,

$NaClO+H_2SO_3\rightarrow NaCl+H_2SO_4$

From the reaction we conclude that the number of electrons exchanged is, 2 electrons.

Molar mass of solute NaClO = 75 g/mole

$\text{Equivalent weight of solute}=\frac{\text{Molar mass of solute}}{\text{number of electrons exchanged}}=\frac{75}{2}=37.5g.eq$

Now we have to calculate the normality of solution.

$\text{Normality}=\frac{15g}{37.5g.eq\times 1L}=0.40eq/L$

Therefore, the correct option is, (a) 0.40