(C) The elevation in boiling point is given by $\Delta T_{b} = K_{b} \cdot m$. Given $\Delta T_{b} = 100.20^{\circ}C - 100^{\circ}C = 0.20^{\circ}C$.

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(C) The elevation in boiling point is given by $\Delta T_{b} = K_{b} \cdot m$. Given $\Delta T_{b} = 100.20^{\circ}C - 100^{\circ}C = 0.20^{\circ}C$. Thus,the molality $m = \frac{\Delta T_{b}}{K_{b}} = \frac{0.20}{0.512} \ mol \ kg^{-1}$. The depression in freezing point is given by $\Delta T_{f} = K_{f} \cdot m$. Substituting the values: $\Delta T_{f} = 1.86 \times \frac{0.20}{0.512} \approx 0.726^{\circ}C$. The freezing point of the solution is $T_{f} = T_{f}^{\circ} - \Delta T_{f} = 0^{\circ}C - 0.726^{\circ}C = -0.726^{\circ}C$.

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$A$ solution of urea (molar mass $60 \ g \ mol^{-1}$) boils at $100.20^{\circ}C$ at atmospheric pressure. If $K_{f}$ and $K_{b}$ for water are $1.86$ and $0.512 \ K \ kg \ mol^{-1}$ respectively,the freezing point of the solution will be:

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A
$-0.654^{\circ}C$
B
$+0.654^{\circ}C$
C
$-0.726^{\circ}C$
D
$+0.726^{\circ}C$

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Elevation in boiling point for $1.5 \, m$ solution of glucose in water is $4 \, K$. The depression in freezing point for $4.5 \, m$ solution of glucose in water is $4 \, K$. The ratio of molal elevation constant to molal depression constant $(K_{b} / K_{f})$ is $....$

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An aqueous solution of urea:

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An aqueous solution freezes at $-2.55 \ ^oC$. What is its boiling point in $^oC$?
[$K_b(H_2O) = 0.52 \ K \ kg \ mol^{-1}, K_f(H_2O) = 1.86 \ K \ kg \ mol^{-1}$]

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The difference between the boiling point and freezing point of an aqueous solution of urea containing $10.0 \ kg$ of water is $100.2372 \ ^oC$. How many grams of urea are dissolved in the solution? (Given: $K_b = 0.513 \ K \ kg \ mol^{-1}$,$K_f = 1.86 \ K \ kg \ mol^{-1}$)

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Column-$I$ (Various solutions) Column-$II$ (Their freezing point)
$a$. $0.1 \ M \ BaCl_2$ solution $p$. $271 \ K$
$b$. $0.1 \ M \ NaCl$ solution $q$. $270 \ K$
$c$. $0.1 \ M \ K_3[Fe(CN)_6]$ solution $s$. $269 \ K$
$d$. $0.1 \ M \ Al_2(SO_4)_3$ solution $r$. $268 \ K$
Given: Freezing point of $0.1 \ M$ sucrose solution $= 272 \ K$ and Freezing point of water $= 273 \ K$. Assume molarity $=$ molality. Which of the following options shows the correct matches?

Column-$I$ (Various solutions)	Column-$II$ (Their freezing point)
$a$. $0.1 \ M \ BaCl_2$ solution	$p$. $271 \ K$
$b$. $0.1 \ M \ NaCl$ solution	$q$. $270 \ K$
$c$. $0.1 \ M \ K_3[Fe(CN)_6]$ solution	$s$. $269 \ K$
$d$. $0.1 \ M \ Al_2(SO_4)_3$ solution	$r$. $268 \ K$

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$A$ solution of urea (molar mass $60 \ g \ mol^{-1}$) boils at $100.20^{\circ}C$ at atmospheric pressure. If $K_{f}$ and $K_{b}$ for water are $1.86$ and $0.512 \ K \ kg \ mol^{-1}$ respectively,the freezing point of the solution will be:

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Elevation in boiling point for $1.5 \, m$ solution of glucose in water is $4 \, K$. The depression in freezing point for $4.5 \, m$ solution of glucose in water is $4 \, K$. The ratio of molal elevation constant to molal depression constant $(K_{b} / K_{f})$ is $....$

An aqueous solution of urea:

An aqueous solution freezes at $-2.55 \ ^oC$. What is its boiling point in $^oC$?[$K_b(H_2O) = 0.52 \ K \ kg \ mol^{-1}, K_f(H_2O) = 1.86 \ K \ kg \ mol^{-1}$]

The difference between the boiling point and freezing point of an aqueous solution of urea containing $10.0 \ kg$ of water is $100.2372 \ ^oC$. How many grams of urea are dissolved in the solution? (Given: $K_b = 0.513 \ K \ kg \ mol^{-1}$,$K_f = 1.86 \ K \ kg \ mol^{-1}$)

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An aqueous solution freezes at $-2.55 \ ^oC$. What is its boiling point in $^oC$?
[$K_b(H_2O) = 0.52 \ K \ kg \ mol^{-1}, K_f(H_2O) = 1.86 \ K \ kg \ mol^{-1}$]