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1 month ago

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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 .

1 answer:

KiRa [2.9K]1 month ago

4 0

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, $1.08\times 10^2g$

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Answer:

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This is a guide to fulfilling the instructions along with essential notes for understanding how to create such graphs:

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In this context, the origin, (0,0), signifies the intersection of the axes at a pressure of 0 mb and a volume of 0.0 milliliters.

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The pressure range along the x-axis is [90, 760 mb], suggesting large divisions of 100 mb, with the farthest right mark at 800 mb. You can then subdivide each 100 mb interval into 10 smaller sections, using small divisions of 10 mb (my example employs 4 sections of 25 mb, but 10 mb is preferable).

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