Question

A solution of glucose (molar mass =180gmol^(-1) ) has been prepared by dissolving 36g of glucose in 500g of water. The freezing point of the solution obtained will be (K_(f) of water =1.86Kkgmol^(-1)) -0.25^(@)C -1.25^(@)C -0.74^(@)C -1.74^(@)C​

Answer 1

Given:

A solution of glucose (molar mass = 180 g mol-') has been prepared by dissolving 36 g of glucose in 500 g of water.

To find:

Freezing point of solution?

Calculation:

First, let's find the DEPRESSION IN FREEZING POINT:

$\rm\Delta T = i \times k_{f} \times m$

• Glucose has Van't Hoff Factor (i.e. 'i') equal to 1.

$\rm \implies\Delta T = 1 \times 1.86 \times \dfrac{moles}{weight \: of \: solvent \: (in \: kg)}$

$\rm \implies\Delta T = 1.86 \times \dfrac{ (\dfrac{36}{180} )}{0.5}$

$\rm \implies\Delta T = 1.86 \times \dfrac{ 0.2}{0.5}$

$\rm \implies\Delta T = {0.744}^{ \circ} \: C$

So, freezing point will be :

$\boxed{ \rm \: T_{freezing} = 0 - 0.744 = - {0.744}^{ \circ} C}$

OPTION C) is CORRECT !

Hope It Helps.