Question

The molarity of HNO₃ in a sample which has density 1.4 g/mL and mass percentage of 63% is :(Molecular weight of HNO₃= 63).

Answer 1

Step 1: Given data

Mass percentage$=63\%$, Density of solution $=1.4\mathrm{g}/\mathrm{ml}$

$63\mathrm{gm}$ of solute is present in $100\mathrm{gm}$ of solution

$\text{Volume}=\frac{\text{Mass}}{\text{density}}$

Step 2 : To find the molarity of a solution

Volume $=\frac{100}{1.4}$

Volume $=71.428\mathrm{ml}$

$\text{Molarity}=\frac{\text{Given wt.}\times 1000}{M.w\times V\text{olume}\left(\mathrm{ml}\right)}$

Molarity $=\frac{63\times 1000}{63\times 71.428}$

Molarity $=14\mathrm{M}$

Hence the molarity of a solution is $14\mathrm{M}$