Question

Calculate the freezing point of a solution containing 15 grams of kcl and 1650.0 grams of water. the molal-freezing-point-depression constant (kf) for water is 1.86 ∘c/m

Answer 1

Answer is: freezing point is -0,226°C.
Answer is: the molal concentration of glucose in this solution is 1,478 m.
m(KCl) = 15 g.
n(KCl) = m(KCl) ÷ M(KCl).
n(KCl) = 15 g ÷ 74,55 g/mol.
n(KCl) = 0,2 mol
m(H₂O) = 1650 g ÷ 1000 g/kg = 1,65 kg.
b = n(KCl) ÷ m(H₂O).
b = 0,2 mol ÷ 1,65 kg = 0,122 m.
Kf(water) = 1,86°C/m.
ΔT = Kf(water) · b(solution).
ΔT = 1,86°C/m · 0,122 m.
ΔT = 0,226°C.

Answer 2

Answer : The freezing point of a solution is $-0.454^oC$

Explanation :  Given,

Molal-freezing-point-depression constant $(K_f)$ for water = $1.86^oC/m$

Mass of KCl (solute) = 15 g

Mass of water (solvent) = 1650.0 g  = 1.650 kg

Molar mass of KCl = 74.5 g/mole

Formula used :  

$\Delta T_f=i\times K_f\times m\\\\T^o-T_s=i\times K_f\times\frac{\text{Mass of KCl}}{\text{Molar mass of KCl}\times \text{Mass of water in Kg}}$

where,

$\Delta T_f$ = change in freezing point

$\Delta T_s$ = freezing point of solution = ?

$\Delta T^o$ = freezing point of water = $0^oC$

i = Van't Hoff factor = 2  (for KCl electrolyte)

$K_f$ = freezing point constant for water = $1.86^oC/m$

m = molality

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

$0^oC-T_s=2\times (1.86^oC/m)\times \frac{15g}{74.5g/mol\times 1.650kg}$

$T_s=-0.454^oC$

Therefore, the freezing point of a solution is $-0.454^oC$