Question

Elevation in the boiling point for 1 molal solution of glucose is 2 K. The depression in the freezing point for 2 molal solution of glucose in the same solvent is 2 K. The relation between ᴷb and ᴷf is :
(A) ᴷb = 1.5 ᴷf (B) ᴷb = ᴷf (C) ᴷb = 0.5 ᴷf (D) ᴷb = 2 ᴷf

Answer 1

hlo bro here is the 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 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} hope it helped plz mark as brainliest