Question

A solution is prepared by dissolving 215 grams of methanol, ch3oh, in 1000. grams of water. what is the freezing point of this solution? [the freezing point depression constant for water is 1.86°c/mole solute in 1000g of water]

Answer 1

Answer : The freezing point of the solution is, 260.503 K

Solution : Given,

Mass of methanol (solute) = 215 g

Mass of water (solvent) = 1000 g = 1 kg       (1 kg = 1000 g)

Freezing depression constant = $1.86^oC/m=1.86Kkg/mole$

Formula used :

$\Delta T_f=K_f\times m\\T^o_f-T_f=K_f\times \frac{w_{solute}}{M_{solute}\times w_{solvent}}$

where,

$T^o_f$ = freezing point of water = $100^oC=273K$

$T_f$ = freezing point of solution

$K_f$ = freezing point constant

$w_{solute}$ = mass of solute

$w_{solvent}$ = mass of solvent

$M_{solute}$ = molar mass of solute

Now put all the given values in the above formula, we get

$273K-T_f=(1.86Kkg/mole)\times \frac{215g}{(32g/mole)\times (1kg)}$

By rearranging the terms, we get the freezing point of solution.

$T_f=260.503K$

Therefore, the freezing point of the solution is, 260.503 K