Question

A 19.57 g mixture of sugar (C12H22O11) and table salt (NaCl) is dissolved in 223 g of water. The freezing point of the solution was measured as -3.85 °C. Calculate the mass percent of sugar in the mixture. A list of Kf values can be found here.

Answer 1

Expression for depression in freezing point can be given as follows:

dT = Kf x m ----------------------------- (1)

Where, Kf is freezing constant = 1.86 K.kg/mol

dT = change in freezing point = 3.85 °C = 3.85 K

m = molality

Molar mass of sugar = 342 g/mol

Molar mass of salt = 58.4 g/mol

Let, y be the grams of sugar, then weight of salt = (19.57 – y)g.

Number of moles of sugar = y/342

Since, NaCl ionizes into two ions, number of mole of solute from NaCl = 2 x (19.57-y)/58.4

Thus, m = [(y/342) + 2(19.57-y)/58.4]/0.223kg

Therefore, equation (1) can be written as follows:

3.85 = (1.86) x [(y/342) + 2(19.57-y)/58.4]/0.223

3.85 / 1.86 = [(y/342) + 2(19.57-y)/58.4]/0.223

2.07 = [(y/342) + 2(19.57-y)/58.4]/0.223

2.07 x 0.223 = [(y/342) + 2(19.57-y)/58.4]

0.4616 = [(y/342) + 2(19.57-y)/58.4]

0.4616 = (y/342) + (39.14/58.4) – (2y/58.4)

0.4616 = (y/342) + 0.6702 – (2y/58.4)

0.4616 - 0.6702 = (y/342)– (2y/58.4)

-0.2086 = y[(1/342) – (2/58.4)]

-0.2086 = y [-0.031326 ]

y = 6.66

Thus, mass percentage of sugar = (6.66g / 19.57g) x 100

mass percentage of sugar = 34.0 %

Therefore, mass percent of sugar in the mixture is 34.0%