Question

Calculate the freezing point of a 2.00 molal solution of the nonelectrolyte glucose. the freezing point constant for water is 1.86 °c/molal.

Answer 1

(1.86 °c / 1 molal)×(2 molal) = 3.72°c, and because the temperature of the freezing point goes down the number has to be negative. So, the final answer is -3.72°c 

Answer 2

Answer : The freezing point of the solution is,
`-3.72^oC`

Explanation : Given,

Molality of the solution = 2.00 m

Formula used :

`\Delta T_f=i* K_f* m\\\\T_f^o-T_f=i* K_f* m`

where,

`\Delta T_f` = change in freezing point

`T_f^o` = temperature of pure water =
`0^oC`

`T_f` = temperature of solution = ?

i = Van't Hoff factor for non-electrolyte solution = 1

`K_f` = freezing point constant =
`1.86^oC/m`

m = molality = 2.00 m

Now put all the given values in this formula, we get:

`0^oC-T_f=1* 1.86^oC/m* 2.00m`

`T_f=-3.72^oC`

Therefore, the freezing point of the solution is,
`-3.72^oC`