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Relative Lowering of Vapour Pressure

A non-volatil...

Question

A non-volatile solute is dissolved in a water. If its degree of association is $50 %$ . What will be the freezing point of solution when its boiling point is ${100.52}^{o}$ C?
$[K_{b} = 0.52$ K kg $m o l^{- 1}$ , $K_{f} = 1.86$ K kg $m o l^{- 1}]$

- 0.52

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{0.52}^{o}

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- {1.86}^{o}

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275.01

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Solution

The correct option is B $- {1.86}^{o}$ C
The molality is calculated from the boiling point elevation
Molality, $m = \frac{Δ T_{b}}{K_{b}} = \frac{0.52 K}{0.52 K K g m o l^{- 1}} = 1 m o l K g^{- 1}$
Freezing point depression= $Δ T_{f}$
$= K_{f} \times m$
$= 1.86 K g m o l^{- 1} \times m o l K g^{- 1}$
$= 1.86 K$
$∴$ Freezing point= $0 - 1.86 = - {1.86}^{o} C$ .

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A non-volatile solute is dissolved in a water. If its degree of association is 50%. What will be the freezing point of solution when its boiling point is 100.52oC?[Kb=0.52K kg mol−1, Kf=1.86K kg mol−1]

The correct option is B −1.86oCThe molality is calculated from the boiling point elevationMolality, m=ΔTbKb=0.52K0.52KKgmol−1=1molKg−1Freezing point depression= ΔTf =Kf×m =1.86Kgmol−1×molKg−1 =1.86K∴ Freezing point= 0−1.86=−1.86oC .

A non-volatile solute is dissolved in a water. If its degree of association is $50 %$ . What will be the freezing point of solution when its boiling point is ${100.52}^{o}$ C?
$[K_{b} = 0.52$ K kg $m o l^{- 1}$ , $K_{f} = 1.86$ K kg $m o l^{- 1}]$