Q64 / 180 · NEET 2019 (Odisha) · Chemistry
The density of $2 \mathrm{M}$ aqueous solution of $\mathrm{NaOH}$ is $1.28 \mathrm{~g} / \mathrm{cm}^3$. The molality of the solution is [Given that molecular mass of $\left.\mathrm{NaOH}=40 \mathrm{~g} \mathrm{~mol}^{-1}\right]$
- A$1.20 \mathrm{~m}$
- B$1.56 \mathrm{~m}$
- Correct answerC$1.67 \mathrm{~m}$
- D$1.32 \mathrm{~m}$
Solution
Given, molarity $=2 \mathrm{M}$ [i.e 2 mole $\mathrm{NaOH}$ in $1 \mathrm{~L}$ solution]
Density $(\mathrm{d})=1.28 \mathrm{~cm}^{-3}$
Molecular weight of $\mathrm{NaOH}=40 \mathrm{gmol}^{-1}$
We know that,
$$
\begin{aligned}
& \text { Density }=\frac{\text { mass of solution }}{\text { volume of solution }} \\
& \therefore \text { Mass of solution }=1.28 \mathrm{~g} \mathrm{~cm}^{-3} \times 1000 \mathrm{~mL}=1280 \mathrm{~g} \\
& \text { Moreover, molarity } \\
& =\frac{\text { number of moles of solute }}{\text { volume of solution }(\text { in } \mathrm{mL})} \times 1000 \\
& \therefore \quad 2=\frac{\text { mass of solute }}{40 \times 1000} \times 1000=80 \mathrm{~g} \\
& \text { Now, mass of solution }=\text { mass of solvent }+ \text { mass of } \\
& \text { solute } \\
& 1280 \mathrm{~g}=\text { mass of solvent }+80 \\
& \therefore \text { Mass of solvent }=1280-80=1200 \mathrm{~g}=1.2 \mathrm{~kg} \\
& \text { Now, molality }=\frac{\text { number of moles of solute }}{\text { mass of solvent }(\text { in kg) }} \\
& =\frac{2}{1.2}=\frac{20}{12}=\frac{5}{3}=1.67 \mathrm{~m} \\
&
\end{aligned}
$$
Alternative method
Molality $(\mathrm{m})=\frac{\mathrm{M}}{1000 \mathrm{~d}-\mathrm{M} \omega \mathrm{t}} \times 1000$
where, $M=$ molarity, $d=$ density of solution
$$
\mathrm{M} \omega \mathrm{t}=\text { molar mass of solute. }
$$
On substing the given values, we get
$$
\begin{aligned}
\text { Molality } & =\frac{2}{1000 \times 1.28-40} \times 1000 \\
& =\frac{2}{1280-40} \times 1000=\frac{2000}{1240}=1.612 \mathrm{~m}
\end{aligned}
$$
Hence, the correct answer is option: 'C'.