Question

A 63% by weight solution of $HN{O}_3$ has a density of 1.5 g/ mL. How much water should be added to 50 mL of this solution so that its normality becomes 1 N?

• A. 980 mL
• B. 750 mL
• C. 700 mL
• D. 400 mL

Answer 1

The correct option is C 700 mL

63% by mass means 63 g of $HN{O}_3$ is present in 100 g of the solution.
Number of moles $=\frac{\text{given mass}}{\text{molar mass}}=\frac{63}{63}=1mol$

Density = mass/volume
so, volume of solution = $\frac{100}{1.5}$ mL
$\text{Molarity}=\frac{\text{moles of solute}}{\text{total volume of solution in mL}}\times 1000$

$=\frac{1}{\frac{100}{1.5}}\times 1000=15M$

Molarity = Normality
(since n-factor is 1 for $HN{O}_3$)
So,
$N_1{V}_1=N_2{V}_2,$ (where $V_2$ is the volume of dilution)
$\Rightarrow (15\times 50)=(1\times V_2)$
$\Rightarrow V_2$ = 750 ml
Hence the amount of water added = (750 – 50) = 700 mL