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Angelina_Jolie [31]
3 years ago
10

A solution is prepared by dissolving 215 grams of methanol, ch3oh, in 1000. grams of water. what is the freezing point of this s

olution? [the freezing point depression constant for water is 1.86°c/mole solute in 1000g of water]
Chemistry
1 answer:
Leya [2.2K]3 years ago
6 0

Answer : The freezing point of the solution is, 260.503 K

Solution : Given,

Mass of methanol (solute) = 215 g

Mass of water (solvent) = 1000 g = 1 kg       (1 kg = 1000 g)

Freezing depression constant = 1.86^oC/m=1.86Kkg/mole

Formula used :

\Delta T_f=K_f\times m\\T^o_f-T_f=K_f\times \frac{w_{solute}}{M_{solute}\times w_{solvent}}

where,

T^o_f = freezing point of water = 100^oC=273K

T_f = freezing point of solution

K_f = freezing point constant

w_{solute} = mass of solute

w_{solvent} = mass of solvent

M_{solute} = molar mass of solute

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

273K-T_f=(1.86Kkg/mole)\times \frac{215g}{(32g/mole)\times (1kg)}

By rearranging the terms, we get the freezing point of solution.

T_f=260.503K

Therefore, the freezing point of the solution is, 260.503 K

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