Q65 / 180 · AIIMS 2019 (26 May) · Chemistry
Calculate molarity of $63 \% \mathrm{w} / \mathrm{w} \mathrm{HNO}_3$ solution if density is $1.4 \mathrm{~g} / \mathrm{mL}$.
- Correct answerA$14 \mathrm{M}$
- B$12 \mathrm{M}$
- C$10 \mathrm{M}$
- D$8\mathrm{M}$
Solution
$63 \%$ nitric acid by mass means that mass of nitric acid $=63 \mathrm{~g}$
Mass of solution $=100 \mathrm{~g}$
Molar mass of $\mathrm{HNO}_3=63 \mathrm{~g} \mathrm{~mol}^{-1}$
Mole of $\mathrm{HNO}_3=\frac{63 \mathrm{~g}}{63 \mathrm{~g} \mathrm{~mol}^{-1}}=1 \mathrm{~mole}$
Density of solution $=1.4 \mathrm{~g} \mathrm{~mL}^{-1}$
$\therefore \quad$ Volume of solution
$=\frac{100 \mathrm{~g}}{1.4 \mathrm{~g} \mathrm{~mL}^{-1}}=71.428 \mathrm{~mL}=0.071428 \mathrm{~L}$
Molarity of solution $=\frac{\text { Moles of solute }}{\text { Volume of solution in } \mathrm{L}}$
$=\frac{1.0}{0.071428} \mathrm{~mol} \mathrm{~L}^{-1}=14 \mathrm{M}$
Hence, the correct answer is option: 'A'.