(B) For acetone as solvent:$\Delta T_{b} = i \times K_{b} \times m$$0.17 = 1 \times 1.7 \times \frac{1.22 / M_{w}}{100 / 1000} \dots (1)$For benzene a

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(B) For acetone as solvent:$\Delta T_{b} = i \times K_{b} \times m$$0.17 = 1 \times 1.7 \times \frac{1.22 / M_{w}}{100 / 1000} \dots (1)$For benzene as solvent (dimerization,$i = 0.5$):$\Delta T_{b} = 0.5 \times 2.6 \times \frac{1.22 / M_{w}}{100 / 1000} \dots (2)$Dividing $(2)$ by $(1)$:$\frac{\Delta T_{b}}{0.17} = \frac{0.5 \times 2.6}{1.7} = \frac{1.3}{1.7}$$\Delta T_{b} = \frac{1.3 \times 0.17}{1.7} = 0.13 \, ^{\circ}C$Given $\Delta T_{b} = x \times 10^{-2} \, ^{\circ}C$,we have $0.13 = x \times 10^{-2}$,so $x = 13$.

Class 12 Chemistry Solutions Mix Examples of Solutions

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$1.22 \, g$ of an organic acid is separately dissolved in $100 \, g$ of benzene $(K_{b}=2.6 \, K \, kg \, mol^{-1})$ and $100 \, g$ of acetone $(K_{b}=1.7 \, K \, kg \, mol^{-1})$. The acid is known to dimerize in benzene but remain as a monomer in acetone. The boiling point of the solution in acetone increases by $0.17^{\circ} C$.
The increase in boiling point of solution in benzene in $^{\circ} C$ is $x \times 10^{-2}$. The value of $x$ is ..... .(Nearest integer)
$[$ Atomic mass : $C=12.0, H=1.0, O=16.0]$

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$12$
B
$13$
C
$10$
D
$11$

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$2 \ \text{moles}$ each of ethylene glycol and glucose are dissolved in $500 \ \text{g}$ of water. The boiling point of the resulting solution is $:$ (Given $:$ Ebullioscopic constant of water $= 0.52 \ \text{K kg mol}^{-1}$) (in $\text{K}$)

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$A$ 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 ........... $^oC$.

DifficultAIIMS 2015

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An aqueous solution freezes at $-0.186\,^\circ C$ ($K_f = 1.86\, K\, kg\, mol^{-1}$; $K_b = 0.512\, K\, kg\, mol^{-1}$). What is the elevation in boiling point?

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The values of $K_b$ and $K_f$ for water are $0.52 \, K \, kg \, mol^{-1}$ and $1.86 \, K \, kg \, mol^{-1}$ respectively. If the solution boils at $0.78 \, K$ above the boiling point of water,then the freezing point of the solution will be ........ $K$.

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Assertion : If one component of a solution obeys Raoult's law over a certain range of composition,the other component will not obey Henry's law in that range.
Reason : Raoult's law is a special case of Henry's law.

MediumAIIMS 2013

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$2 \ \text{moles}$ each of ethylene glycol and glucose are dissolved in $500 \ \text{g}$ of water. The boiling point of the resulting solution is $:$ (Given $:$ Ebullioscopic constant of water $= 0.52 \ \text{K kg mol}^{-1}$) (in $\text{K}$)

$A$ 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 ........... $^oC$.

An aqueous solution freezes at $-0.186\,^\circ C$ ($K_f = 1.86\, K\, kg\, mol^{-1}$; $K_b = 0.512\, K\, kg\, mol^{-1}$). What is the elevation in boiling point?

The values of $K_b$ and $K_f$ for water are $0.52 \, K \, kg \, mol^{-1}$ and $1.86 \, K \, kg \, mol^{-1}$ respectively. If the solution boils at $0.78 \, K$ above the boiling point of water,then the freezing point of the solution will be ........ $K$.

Assertion : If one component of a solution obeys Raoult's law over a certain range of composition,the other component will not obey Henry's law in that range.Reason : Raoult's law is a special case of Henry's law.

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Assertion : If one component of a solution obeys Raoult's law over a certain range of composition,the other component will not obey Henry's law in that range.
Reason : Raoult's law is a special case of Henry's law.