A solution of urea (molar mass 60 , g , mol^{ | vedclass

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(B) Given: Boiling point of solution = $100.18 \, ^oC$. Boiling point of pure water = $100 \, ^oC$. Elevation in boiling point,$\Delta T_b = 100.18 -

Vedclass · Accepted Answer

$-0.65$

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(B) Given: Boiling point of solution = $100.18 \, ^oC$. Boiling point of pure water = $100 \, ^oC$. Elevation in boiling point,$\Delta T_b = 100.18 - 100 = 0.18 \, ^oC$. We know that $\Delta T_b = K_b \cdot m$ and $\Delta T_f = K_f \cdot m$. Therefore,$\frac{\Delta T_f}{\Delta T_b} = \frac{K_f}{K_b}$. $\Delta T_f = \Delta T_b 	imes \frac{K_f}{K_b} = 0.18 	imes \frac{1.86}{0.512} \approx 0.654 \, ^oC$. Since $\Delta T_f = T_f^{\circ} - T_f$,where $T_f^{\circ} = 0 \, ^oC$ for water: $0.654 = 0 - T_f$. $T_f = -0.654 \, ^oC$. Rounding to two decimal places,the freezing point is $-0.65 \, ^oC$.

$A$ solution of urea (molar mass $60 \, g \, mol^{-1}$) boils at $100.18 \, ^oC$ at atmospheric pressure. If $K_f$ and $K_b$ for water are $1.86$ and $0.512 \, K \, kg \, mol^{-1}$ respectively,the above solution will freeze at ........... $^oC$.

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