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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 K_band K_fis:
Option 1)
K_b= K_f
Option 2)
K_b=1.5 K_f
Option 3)
K_b=0.5 K_f
Option 4)
K_b=2 K_f

Answers (1)

best_answer

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 T_band T_f .

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

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

$K_{b} = 2K_{f}$

Option 1)

K_b= K_f

Option 2)

K_b=1.5 K_f

Option 3)

K_b=0.5 K_f

Option 4)
K_b=2 K_f

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