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Tell me? Elevation in the boiling point for 1 molal solution og glucose is 2 K .the depression inthe freezing point for 2molal solution of glucose in the same solvent is 2 ,K.the relatin between Kb and Kfis:

Elevation in the boiling point for 1 molal solution og glucose is 2 K .the depression inthe freezing point for 2molal solution of glucose in the same solvent is 2  ,K.the relatin between Kb and Kfis:

  • Option 1)

    Kb = Kf

  • Option 2)

    Kb =1.5 Kf

  • Option 3)

    Kb =0.5 Kf

  • Option 4)

    Kb =2 Kf

Answers (1)
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Mathematical Expression -

\Delta T_{b}= K_{b}\: m

Unis of K_{b}=\frac{K-K_{g}}{mole}

\Delta T_{b}= Elevation\: in \: boiling\: point
 

- wherein

K_{b}= Boiling \: point \: elevation \: constant

m= molality

 

 

Mathematical Expression of Depression in Freezing point -

\Delta T_{f}= K_{f}\: m
 

- wherein

m = molarity of solvent 

K_{f} = cryoscopic  constant

    molal depress const

Units = \frac{K-K_{g}}{mole}

 

As we have learned

The change in Tb and Tf .

\frac{\Delta T_{b}}{\Delta T_{f}} = \frac{i \times m \times K_{b}}{i \times m \times K_{f}}

\frac{2}{2} = \frac{1 \times 1 \times K_{b}}{1 \times 2 \times K_{f}}

K_{b} = 2K_{f}


Option 1)

Kb = Kf

Option 2)

Kb =1.5 Kf

Option 3)

Kb =0.5 Kf

Option 4)

Kb =2 Kf

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