Question

Which of the following strengths of solution is equivalent to $N$ solution of sodium carbonate?
A.$106{\text{gm per 100ml}}$
B.\[{\text{53gm per 100ml}}\]
C.$10.6{\text{gm per 100ml}}$
D.${\text{5}}{\text{.3gm per 100ml}}\,$

Answer 1

Hint: Strength of the solution is the amount/quantity of solute in the solution.
Normality is defined as the number of equivalents of solute dissolved per litre of solution. It is calculated by dividing the molecular weight of solute by the number of equivalents per mole of solute.

Complete step by step answer:
We know that the formula to calculate the normality of the solution is equal to ${\text{Normality = }}\dfrac{{{\text{weight}} \times 1000}}{{{\text{gram equivalent}} \times {\text{ volume in ml}}}}$
 Gram equivalent for sodium carbonate is $ = \dfrac{{23 \times 2 + 12 + 16 \times 3}}{2}$$ = 53$
For comparing the strength we have to calculate the normality of each option given in the question.
So for first option which is $106{\text{gm per 100ml}}$
For this apply the formula of normality, $N = \dfrac{{106 \times 1000}}{{53 \times 100}} = 20$
Applying the same for option (ii) which is \[{\text{53gm per 100ml}}\] then normality will be, $N = \dfrac{{53 \times 1000}}{{53 \times 100}} = 10$
Applying the same for option (iii) which is $10.6{\text{gm per 100ml}}$ then normality will be,
$N = \dfrac{{10.6 \times 1000}}{{53 \times 100}} = 2$
Applying the same for option (iii) which is ${\text{5}}{\text{.3gm per 100ml}}\,$then normality will be,
$N = \dfrac{{5.3 \times 1000}}{{53 \times 100}} = 1$
Hence, the required answer will be ${\text{5}}{\text{.3gm per 100ml}}\,$as it has $1N$ as its solution strength.

Note:
Apply the formula of normality carefully, as it contains gram equivalent so we have to divide it by its n-factor. n-factor is defined as the number of ${H^ + }$ions replaced by $1$ mole of acid in a reaction.