The elevation in the boiling point of aqueous urea solution is $0.104 \mathrm{~K}$. What is its $\Delta…

The elevation in the boiling point of aqueous urea solution is $0.104 \mathrm{~K}$. What is its $\Delta \mathrm{T}_{\mathrm{f}}$ (in $\mathrm{K}$ ) value? (for Water $\mathrm{K}_{\mathrm{b}}=0.52 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}, \mathrm{~K}_{\mathrm{f}}=1.86 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}$ )
  1. 0.0186
  2. 0.186
  3. 0.372
  4. 0.0372

Solution

$\Delta \mathrm{T}_{\mathrm{b}}=0.104 \mathrm{~K}, \mathrm{~K}_{\mathrm{b}}=0.52 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}, \mathrm{~K}_{\mathrm{f}}=1.86 \mathrm{~K}$ $\mathrm{kg} \mathrm{mol}^{-1}$ Thus, $\Delta \mathrm{T}_{\mathrm{b}}=\mathrm{m} \mathrm{K}_{\mathrm{b}} \Rightarrow \mathrm{m}=\frac{\Delta \mathrm{T}_{\mathrm{b}}}{\mathrm{K}_{\mathrm{b}}}=\frac{0.104}{0.52}$ $=0.2 \mathrm{~mol} \mathrm{~kg}^{-1}$ $\begin{aligned} & \Rightarrow \Delta \mathrm{T}_{\mathrm{f}}=\mathrm{m} \mathrm{K}_{\mathrm{f}}=0.2 \times 1.86 \\ & =0.372 \mathrm{~K}\end{aligned}$

Asked in: AP EAMCET 2023 (16 May Shift 1)

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