Question

Calculate change in freezing point of water if 0.01 more Potassium Sulphate is dissolved in 1 kg of water kf
for water is equal to 1.86 kg mole inverse

Answer 1

Answer : The change in freezing point of water is, 0.0186 K

Solution : Given,

Moles of potassium sulfate(solute) = 0.01 moles

Mass of water(solvent) = 1 kg

$k_f=1.86Kkg/mole$

Formula used for lowering in freezing point is,

$\Delta T_f=k_f\times m$

or,

$\Delta T_f=k_f\times \frac{\text{ Moles of solute}}{\text{ Mass of solvent in kg}}$

where,

$T_f$ = change in freezing point

$k_f$ = freezing point constant

m = molality

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

$\Delta T_f=(1.86Kkg/mole)\times \frac{0.01mole}{1kg}=0.0186K$

Therefore, the change in freezing point of water is, 0.0186 K