Q7 / 90 · JEE Main 2024 (06 Apr Shift 1) · Chemistry

The density of ' \(x\) ' \(\mathrm{M}\) solution (' \(X\) ' molar) of \(\mathrm{NaOH}\) is \(1.12 \mathrm{~g} \mathrm{~mL}^{-1}\), while in molality, the concentration of the solution is \(3 \mathrm{~m}(3 \mathrm{molal})\). Then \(x\) is (Given : Molar mass of \(\mathrm{NaOH}\) is \(40 \mathrm{~g} / \mathrm{mol}\) )

  • A
    3.5
  • B
    3.8
  • C
    2.8
  • Correct answer
    D
    3.0

Solution

$\begin{aligned} & \text { Molality }=\frac{1000 \times \mathrm{M}}{1000 \times \mathrm{d}-\mathrm{M} \times(\mathrm{Mw})_{\text {solute }}} \\ & 3=\frac{1000 \times \mathrm{x}}{1000 \times 1.12-(\mathrm{x} \times 40)} \\ & \mathrm{x}=3\end{aligned}$

Hence, the correct answer is option: 'D'.

Relevant Questions