Question

What vol of 0.1N HNO3 solution can be prepared from 6.3g of HNO3

Answer 1

Answer: 1 Litre of $HNO_3$ solution of 0.1 N.

Solution:

$\text{Normality}=\frac{\text{given weight}}{\text{equivalent weight}}\times \frac{1}{\text{volume in litres}}$

Given weight of nitric acid: 6.3 g

Equivalent weight of nitric acid : $\frac{\text{molecular mass}}{\text{number of displacable hydrogens}}=\frac{63 g}{1}=63 \text{gram equivalent}$

Volume prepared from 6.3 g of nitric acid:

$0.1N=\frac{6.3 g}{63}\times \frac{1}{\text{Volume}}$

$V=1 L$

1 Litre of 0.1N nitric acid solution will be prepared from 6.3g of nitric acid.

Answer 2

Answer: The volume of solution is 1 L.

Explanation:

Equivalent mass of an acid is defined as the ratio of its molecular mass and basicity.

$\text{Equivalent mass of acid}=\frac{\text{Molar mass}}{\text{Basicity}}$

Molar mass of nitric acid = 63 g/mol

Basicity of nitric acid = 1  (one replacable hydrogen ion)

Putting values in above equation, we get:

$\text{Equivalent mass of }HNO_3=\frac{63}{1}=63g/eq.$

To calculate the volume of solution for given value of normality, we use the equation:

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

We are given:

Normality of solution = 0.1 eq/L

Weight of nitric acid = 6.3 g

Equivalent weight of nitric acid = 63 g/eq.

Volume of solution = ? L

Putting values in above equation, we get:

$0.1eq./L=\frac{6.3g}{63g/eq.\times \text{Volume of solution}}\\\\\text{Volume of solution}=1L$

Hence, the volume of solution is 1 L.