Assuming complete dissociation of the solute, how many grams of KNO3 must be added to 275 mL of water to produce a solution that

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3 months ago

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freezes at −14.5 ∘C? The freezing point for pure water is 0.0 ∘C and Kf is equal to 1.86 ∘C/m .

1 answer:

KiRa [2.9K] · Answer 1 · 2026-02-27T13:41:17+03:00

Answer: The required mass of KNO₃ is, $1.08\times 10^2g$

Explanation: Provided,

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

Amount of water = 275 mL

Molar mass of KNO₃ = 101.1 g/mole

First, we need to determine the mass of the water.

$\text{Mass of water}=\text{Density of water}\times \text{Volume of water}$

Density of water = 1.00 g/mL

$\text{Mass of water}=1.00g/mL\times 275mL=275g=0.275kg$

Next, we will figure out the mass of KNO₃

Formula utilized:

$\Delta T_f=i\times K_f\times m\\\\T^o-T_s=i\times K_f\times\frac{\text{Mass of }KNO_3}{\text{Molar mass of }KNO_3\times \text{Mass of water in Kg}}$

where,

$\Delta T_f$ = change in freezing point

$\Delta T_s$ = freezing point of solution = $-14.5^oC$

$\Delta T^o$ = freezing point of water = $0.0^oC$

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

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

m = molality

Substituting all known values into this equation yields

$0.0^oC-(-14.5^oC)=2\times (1.86^oC/m)\times \frac{\text{Mass of }KNO_3}{101.1g/mol\times 0.275kg}$

$\text{Mass of }KNO_3=108.369g=1.08\times 10^2g$

Consequently, the mass of KNO₃ that must be added is, $1.08\times 10^2g$