(D) For glucose,the van't Hoff factor $i = 1$.The elevation in boiling point is given by $\Delta T_{b} = K_{b} \cdot m_{1} = 4 \, K$,where $m_{1} = 1.

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(D) For glucose,the van't Hoff factor $i = 1$.The elevation in boiling point is given by $\Delta T_{b} = K_{b} \cdot m_{1} = 4 \, K$,where $m_{1} = 1.5 \, m$.So,$K_{b} \cdot 1.5 = 4 \implies K_{b} = \frac{4}{1.5}$.The depression in freezing point is given by $\Delta T_{f} = K_{f} \cdot m_{2} = 4 \, K$,where $m_{2} = 4.5 \, m$.So,$K_{f} \cdot 4.5 = 4 \implies K_{f} = \frac{4}{4.5}$.The ratio $\frac{K_{b}}{K_{f}} = \frac{4 / 1.5}{4 / 4.5} = \frac{4.5}{1.5} = 3$.

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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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$4$
B
$1$
C
$2$
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$3$

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At $40\,^oC$,the vapour pressure (in torr) of a solution of methyl alcohol $(A)$ and ethyl alcohol $(B)$ is represented by: $P_s = 120 X_A + 138$,where $X_A$ is the mole fraction of methyl alcohol. The values of $\lim_{X_A \to 0} (P_B^0)$ and $\lim_{X_B \to 0} (P_A^0)$ are:

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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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$A$ liquid freezes at $7^{\circ} C$ and boils at $77^{\circ} C$. If $K_{f}$ and $K_{b}$ values for the liquid are $5.6$ and $2.5 \ ^{\circ} C \ kg \ mol^{-1}$ respectively,then the ratio of latent heat of vaporisation of the liquid to the latent heat of fusion is $:-$

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Identify the correct statement from the following properties.

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An aqueous solution of a solute $AB$ has a normal boiling point of $101.08\,^{\circ}C$ and a normal freezing point of $-1.80\,^{\circ}C$. Hence,$AB$:
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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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At $40\,^oC$,the vapour pressure (in torr) of a solution of methyl alcohol $(A)$ and ethyl alcohol $(B)$ is represented by: $P_s = 120 X_A + 138$,where $X_A$ is the mole fraction of methyl alcohol. The values of $\lim_{X_A \to 0} (P_B^0)$ and $\lim_{X_B \to 0} (P_A^0)$ are:

$2 \ \text{moles}$ 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 $\text{K}$)

$A$ liquid freezes at $7^{\circ} C$ and boils at $77^{\circ} C$. If $K_{f}$ and $K_{b}$ values for the liquid are $5.6$ and $2.5 \ ^{\circ} C \ kg \ mol^{-1}$ respectively,then the ratio of latent heat of vaporisation of the liquid to the latent heat of fusion is $:-$

Identify the correct statement from the following properties.

An aqueous solution of a solute $AB$ has a normal boiling point of $101.08\,^{\circ}C$ and a normal freezing point of $-1.80\,^{\circ}C$. Hence,$AB$:Given: $AB$ is $100\%$ ionised at the boiling point of the solution,$(K_b / K_f)_{\text{water}} = 0.3$.

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An aqueous solution of a solute $AB$ has a normal boiling point of $101.08\,^{\circ}C$ and a normal freezing point of $-1.80\,^{\circ}C$. Hence,$AB$:
Given: $AB$ is $100\%$ ionised at the boiling point of the solution,$(K_b / K_f)_{\text{water}} = 0.3$.