Assuming complete dissociation of the solute, how many grams of KNO3 must be added to 275 mL of water to produce a solution that freezes at −14.5 ∘C? The freezing point for pure water is 0.0 ∘C and Kf is equal to 1.86 ∘C/m .

Answer : The mass of KNO₃ added must be,

Explanation : Given,

Molal-freezing-point-depression constant for water =

Volume of water = 275 mL

Molar mass of KNO₃ = 101.1 g/mole

First we have to calculate the mass of water.

Density of water = 1.00 g/mL

Now we have to calculate the mass of KNO₃

Formula used :  

where,

= change in freezing point

= freezing point of solution =

= freezing point of water =

i = Van't Hoff factor = 2  (for KNO₃ electrolyte)

= freezing point constant for water =

m = molality

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

Therefore, the mass of KNO₃ added must be,

i think it is true since you can test the answer and can use observations to complete it.

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