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. Express your answer to three significant figures and include the appropriate units.

[tex]KNO_{3}[/tex] is a strong electrolyte
For an electrolyte dissolved in a solution- [tex]\Delta T_{f}=i.K_{f}.C[/tex]
[tex]\Delta T_{f}[/tex] is depression in freezing point of solution, i is van't hoff factor (equal to number of ions produce from dissociation of 1 molecule of electrolyte) and C is molality of solution
Molar mass of [tex]KNO_{3}[/tex] is 101.1 g/mol and density of water is 1 g/mL
So, mass of 275 mL of water is 275 g
We know, molality = (number of moles of solute)/(mass of solvent in kg)
If W g of [tex]KNO_{3}[/tex] must be added then [tex]C=\frac{\frac{W}{101.1}}{0.275}(m)[/tex]
For [tex]KNO_{3}[/tex], i = 2 ([tex]KNO_{3}\rightarrow K^{+}+NO_{3}^{-}[/tex])
Here [tex]\Delta T_{f}=(0.0)-(-14.5)^{0}\textrm{C}=14.5^{0}\textrm{C}[/tex]

So, [tex]14.5=2\times 1.86\times \frac{\frac{W}{101.1}}{0.275}[/tex]

or, W = 108

So, 108 g of [tex]KNO_{3}[/tex] must be added

pstnonsonjoku pstnonsonjoku · Answer 2 · 2021-11-07T16:23:04+01:00

The mass of KNO3 required is 108.3 g of KNO3.

From the question, we have the following information;

K = 1.86 ∘C/m

Mass of solution = 275 g or 0.275 Kg

Freezing point of solution = −14.5 ∘C

Freezing point of pure water = 0.0∘C

We know that;

ΔT = Freezing point of solvent- Freezing point of solution

ΔT = 0∘C - ( −14.5 ∘C)

ΔT = 14.5 ∘C

Then;

ΔT = K m i

K = freezing constant

m = molality

i = Van't Hoff factor

Since KNO3 produces two particles, i = 2

Making the molality the subject of the formula;

m = ΔT/K i

m = 14.5 ∘C/ 1.86 ∘C/m × 2

m = 3.898 m

But molality = number of moles/mass of solution in kilograms

number of moles of solute = molality × mass of solution in kilograms

number of moles of solute= 3.898 m × 0.275 Kg

number of moles of solute= 1.072 moles

Number of moles = mass/molar mass

Molar mass of KNO3 = 101 g/mol

Mass = Number of moles × 101 g/mol

Mass = 1.072 moles × 101 g/mol

Mass = 108.3 g of KNO3

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