Question

the freezing point of 0.05 m solution of glucose in water is KF is equal to 1.86 degree Celsius per m​

Answer 1

$\huge\bold{HEY:)-}$

Tf = Kf  x m = 1.86 x 0.05 = 0.093°C

Tf = 0 – 0.093 =  -0.093°C

Answer 2

Answer:

The freezing point of 0.05 m solution of glucose in water is 0.093°C.

Explanation:

We will use the following equation to solve this question which is given as,

$\mathrm{\Delta T_{f}=k_{f}\times m}$      (1)

Where,

$\mathrm{\Delta T_{f}}$=change in freezing point of the solution

$\mathrm{k_{f}}$=molal depression constant

m=molality of the solution

From the question we have,

The molality of the solution=0.05m

The molal depression is constant=1.86°C/m

By inserting the values in equation (1) we get;

$\mathrm{\Delta T_{f}=1.86\times 0.05}$

$\mathrm{\Delta T_{f}=0.093^{\circ}C}$         (2)

And depression at the freezing point is calculated as,

$\mathrm{\Delta T_{f}=T_1-T_2}$        (3)

T₁=freezing point of the solvent=0°C

T₂=freezing point of solution

By inserting values in equation (3) we get;

$\mathrm{0.093^{\circ}=0^{\circ}-T_2}$

$\mathrm{T_2=0.093^{\circ}C}$

Hence, the freezing point of 0.05 m solution of glucose in water is 0.093°C.

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