Question

Calculate normality of 2.1 %( w/V) sulphuric acid solution.
(a) 2.14N
(b) 4.28N
(c) 0.428N
(d) 0.214N

Answer 1

Hint: By knowing the equivalent mass of the substances, we can calculate the normality of the solution. When a substance is given in percentage, we can take the percent in mass in 100mL solution. The equation of normality is given as:

$\text{Normality}=\frac{\text{Equivalents}\times 100}{\text{Volume of Solution}}$

 Complete step by step answer:
Normality is equal to the gram equivalent weight per litre of solution. It is a measure of concentration. Gram equivalent weight is the measure of reactive capacity of a molecule. Normality can also be defined as number of mole equivalents or number of gram of solute present in 1 litre of a solution. Equivalent is the number of moles of reactive units in a compound.
We can consider 2.1% of sulphuric acid as 2.1g of sulphuric acid in 100mL solution as it is given in the question that the percentage is weight by volume.
So, using this information we can find the gram equivalent mass. Gram equivalent mass of a substance is molar mass divided by the valency of the substance. For sulphuric acid it will be,
Gram equivalent mass=$\frac{\text{Molar Mass}}{\text{Valency}}=\frac{98}{2}=49$
Now number of gram equivalent mass is the given weight of sulphuric acid = 2.1 divided by the gram equivalent mass = 49 and it is written as:
Number of gram equivalent mass = $\frac{2.1}{49}$
Let us consider the equation for normality,
 $\text{Normality}=\frac{\text{Equivalents}\times 100}{\text{Volume of Solution}}$
Here, equivalents = $\frac{2.1}{49}$ and the total volume of the solution we know is 100mL.
Substituting these values we get,
Normality = $\frac{\frac{2.1}{49}\times 1000}{100}$ = 0.428N.
Thus, the normality of 2.1% of sulphuric acid is 0.428N.

So, the correct answer is “Option C”.

Note:
There might be confusion between normality and molarity. Normality depends on gram equivalent per litre of solution and molarity depends on the number of moles of solute per litre of solution.