Q22 / 160 · AP EAMCET 2023 (17 May Shift 1) · Chemistry

The molality of solution, when $18 \mathrm{~g}$ of glucose is added to the $18 \mathrm{~g}_{\text {of } \mathrm{H}_2 \mathrm{O} \text { is }}$

  • A
    $0.55 \mathrm{~m}$
  • B
    $2.55 \mathrm{~m}$
  • Correct answer
    C
    $5.55 \mathrm{~m}$
  • D
    $55.5 \mathrm{~m}$

Solution

$\begin{aligned} & \text {} \mathrm{m}_2=18 \mathrm{~g}, \mathrm{M}_2=180 \mathrm{~g} \mathrm{~mol}^{-1}, \mathrm{~m}_1=18 \mathrm{~g}, \mathrm{M}_1=18 \mathrm{~g} \\ & \begin{aligned} \Rightarrow \mathrm{n}_2=\frac{\mathrm{m}_2}{\mathrm{M}_2} & =\frac{18}{180}=0.1 \mathrm{~mol} \\ \mathrm{M}_1=18.0 \mathrm{~g} & =18.0 \times 10^{-3} \mathrm{~kg} \\ \Rightarrow \text { molality } & =\frac{\text { Number of moles of solute }}{\text { Mass of solvent in } \mathrm{kg}}=\frac{0.1}{18.0 \times 10^{-3}} \\ & =5.55 \mathrm{~mol} \mathrm{~kg}^{-1} .\end{aligned}\end{aligned}$

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

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