Question

Normality of 10(W/V) ${H}_{2}S{O}_{4}$ solution is nearly:
a) 0.1
b) 0.2
c) 0.5
d) 2

Answer 1

Hint: Normality is an expression that is used to calculate or measure the concentration of a solution. It relates the amount of solute to the total volume of the solution, just like molarity.

Complete step by step answer:
Normality is the concentration equal to the gram equivalent weight per litre of solution. The gram equivalent weight is defined as the measure of reactive capacity of a molecule. It is abbreviated as 'N'. The formula of normality is given as follow:

$Normality\quad =\quad \dfrac { No.\quad of\quad gram\quad equivalents }{ Volume\quad of\quad solution\quad in\quad litres }$, or

$Normality\quad =\quad \dfrac { weight\quad (g) }{ equivalent\quad weight\quad (g/eq) } \quad \times \quad \dfrac { 1000 }{ volume\quad of\quad solvent\quad (ml) }$

Now according to the question we know that ${H}_{2}S{O}_{4}$ is a dibasic acid i.e., it has replaceable hydrogen atoms

Thus, $equivalent\quad weight\quad =\quad \dfrac { Molecular\quad weight }{ 2 }$

$\implies equivalent\quad weight\quad =\quad \dfrac { 98.079 }{ 2 }$

$\implies equivalent\quad weight\quad =\quad 49.0395\quad g$

Now, from the question, 10% W/V = 10g ${H}_{2}S{O}_{4}$ in 100 ml water. Therefore, weight(g)=10g, volume of solvent = 100ml and equivalent weight = 49.0395 g. Substituting these in equation (1), we get

$Normality\quad =\quad \dfrac { 10 }{ 49.0395 } \quad \times \quad \dfrac { 1000 }{ 100 } = 2.0369$

$\implies Normality = 2 N$

Hence, option (d) is the correct answer.

Note: Do not confuse between molarity (M) and normality (N). Molarity (M) is defined as the number of moles of solute per litre of solution. Normality (N) is defined as the number of mole equivalents per litre of solution.