The freezing point of a solution made by dissolving 19.44 g of a certain sugar in 225 g of water is -4.00 °C. What is the molar mass of the sugar? The \( K_f \) for water is 1.86 °C/m.

A. 40.2 g/mol
B. 204 g/mol
C. 17.7 g/mol
D. 38.1 g/mol

Question

High School

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Answer :

SinghShube SinghShube · Answer 1 · 2025-01-19T13:27:00-05:00

The molar mass of the sugar is 204 g/mol. So, out of the given options b is correct.

To determine the molar mass of the sugar, we can use the formula for freezing point depression:

[tex]\triangle T_f = K_f m[/tex]

where ΔTf is the freezing point depression, Kf is the cryoscopic constant for water (1.86 °C/m), and m is the molality of the solution.

We are given the freezing point depression (ΔTf) as 4.00 °C and the mass of water (225 g). We need to calculate the molality (m) of the solution.

First, we calculate the moles of sugar:

moles of sugar = mass of sugar / molar mass

Next, we calculate the molality:

molality (m) = moles of sugar / mass of water (in kg)

So, mass of sugar = 19.44 g

mass of water = 225 g = 0.225 kg

So, moles of sugar = 19.44 g / molar mass

Hence, molality (m) = (19.44 g / molar mass) / 0.225 kg

Substituting the given values into the freezing point depression equation:

4.00 °C = (1.86 °C/m) × [(19.44 g / molar mass) / 0.225 kg]

So, molar mass = (1.86 °C/m) × [(19.44 g) / (0.225 kg × 4.00 °C)]

molar mass = 204 g/mol

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