(A) Given: $\Delta T_{b} = 0.01 \ K$,molality $(m) = 0.01 \ m$,and $K_{b} = 0.52 \ K \ kg \ mol^{-1}$.We use the formula for elevation in boiling poin

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(A) Given: $\Delta T_{b} = 0.01 \ K$,molality $(m) = 0.01 \ m$,and $K_{b} = 0.52 \ K \ kg \ mol^{-1}$.We use the formula for elevation in boiling point: $\Delta T_{b} = i \times K_{b} \times m$.Rearranging to solve for the Van't Hoff factor $(i)$: $i = \frac{\Delta T_{b}}{K_{b} \times m}$.Substituting the values: $i = \frac{0.01}{0.52 \times 0.01} = \frac{1}{0.52} \approx 1.92$.Thus,the Van't Hoff factor is $1.92$.

Class 12 Chemistry Solutions Colligative properties of electrolyte

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The $\Delta T_{b}$ value for $0.01 \ m$ $KCl$ solution is $0.01 \ K$. What is the Van't Hoff factor? $(K_{b}$ for water $= 0.52 \ K \ kg \ mol^{-1})$

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A
$1.92$
B
$1.72$
C
$0.96$
D
$0.86$

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Calculate the approximate $\Delta T_b$ (in $K$) for $0.001 \ m$ $KCl$ solution, if its van't-Hoff factor is $1.98$ [$K_b$ of water is $0.52 \ K \ kg \ mol^{-1}$].

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The $\Delta T_{b}$ value for $0.01 \ m$ $KCl$ solution is $0.01 \ K$. What is the Van't Hoff factor? $(K_{b}$ for water $= 0.52 \ K \ kg \ mol^{-1})$

Similar Questions

Which of the following will have the highest depression in freezing point when their concentration is $0.1 \ M$?

Which of the following aqueous solutions has the highest boiling point?

The molal elevation constant of water is $0.52 \, ^{\circ}C \, kg \, mol^{-1}$. The boiling point of $1.0 \, m$ aqueous $KCl$ solution (assuming complete dissociation of $KCl$) should be $......... \, ^{\circ}C$.

Calculate the approximate $\Delta T_b$ (in $K$) for $0.001 \ m$ $KCl$ solution, if its van't-Hoff factor is $1.98$ [$K_b$ of water is $0.52 \ K \ kg \ mol^{-1}$].

The correct order of boiling point of the given aqueous solutions is:

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