The solution containing $3 \mathrm{~g}$ urea (molar mass 60 ) per $\mathrm{dm}^3$ of water and another…

The solution containing $3 \mathrm{~g}$ urea (molar mass 60 ) per $\mathrm{dm}^3$ of water and another solution containing $4.5 \mathrm{~g}$ of solute A per $\mathrm{dm}^3$ boils at same temperature, then what is molar mass of A?
  1. $54 \mathrm{~g} \mathrm{~mol}^{-1}$
  2. $180 \mathrm{~g} \mathrm{~mol}^{-1}$
  3. $120 \mathrm{~g} \mathrm{~mol}^{-1}$
  4. $90 \mathrm{~g} \mathrm{~mol}^{-1}$

Solution

$\Delta \mathrm{T}_{\mathrm{b}}=\mathrm{K}_{\mathrm{b}} \cdot \mathrm{m}$ $\text { molality }(\mathrm{m})=\frac{\text { moles of solute }}{\text { mass of water }(\mathrm{kg})}$ 1L $\mathrm{H}_2 \mathrm{O}=1 \mathrm{dm}^3 \mathrm{H}_2 \mathrm{O}=1 \mathrm{~kg} \mathrm{H}_2 \mathrm{O}$ Density of water $=1 \mathrm{~kg} / \mathrm{L}$ Boiling points are same for two solutions, $\begin{aligned} & \Delta \mathrm{T}_{\mathrm{b}_1}=\Delta \mathrm{T}_{\mathrm{b}_2} \\ & \mathrm{~m}_1=\mathrm{m}_2 \end{aligned}$ $\begin{aligned} & \frac{3 / 60}{1}=\frac{4.5 / \mathrm{M}}{1} \\ & \mathrm{M}=\frac{60 \times 4.5}{3} \\ & =90 \mathrm{~g} / \mathrm{mol} \end{aligned}$

Asked in: MHT CET 2021 (24 Sep Shift 2)

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