Question

For the preparation of a litre of \[\dfrac{N}{{10}}\] solution of ${H_2}S{O_4}$, we need:
A. $9.8$ g
B. $4.9$ g
C. $10$ g
D. $98$ g

Answer 1

Hint: Normality is the measure of the concentration of solute in solution. It is also defined as the number of mole equivalents per liter of solution. There is a relation in Normality (N), number of mole equivalents (n), litres of solution(V) that is
Normality = number of mole equivalents/1 L of solution

Complete step by step solution:

To find out molar concentration, $M = \dfrac{n}{V}$ then $M = \dfrac{{1mole}}{{1L}}$
We get $M = 1$ and this will help in approaching the problem .
We have a N/10 solution i.e. 0.1 Normal solution.
We also know that
Normality= Molarity $\times$ Basicity (relation between normality and molarity)
And the basicity of ${H_2}S{O_4}$ is 2 because it is a strong acid and gives all ${H^ + }$ ions when ionized.
1 Molar solution of ${H_2}S{O_4}$ is equal to a 2 N solution as we have already found out the molarity of solution.
Molecular weight of ${H_2}S{O_4}$$ = 2 \times 1 + 32 \times 1 + 16 \times 4 = 98$
It means 1 M solution we would require $98g = 2N$ solution.
1 N solution of 1 litre would require $\dfrac{{98}}{2} = 49$ g
for a 0.1 N solution $ = 4.9g$

Note: In this question we are given a solution of ${H_2}S{O_4}$ who’s normality is \[\dfrac{N}{{10}}\]. So first of all we should be aware about the concept of normality and molarity and in the process of solving this problem we would find out molecular weight of ${H_2}S{O_4}$ then by using $M = \dfrac{n}{V}$ we will find out molarity and here $M = 1$ as it is mention in the problem that it is 1 L of solution so $V = 1$ and $n = 1$ because when no of moles are not mentioned we always take it 1 mole, then comparing the molarity with normality we would find out that \[\dfrac{N}{{10}}\] solution of ${H_2}S{O_4}$ will require 4.9 g of ${H_2}S{O_4}$ as 1 N solution required 49 g of ${H_2}S{O_4}$.