How is this solved Question 12 A studentrpreparced a sugar

Question #: 12 A studentrpreparced a sugar solution containig 6 C046 of wgr (Cul)in 250 al.of ward he/she produced the following freezing point data: Freezing Point of Pure Water Freezing Point of Sugar Solution What is the Kr of water based on the student\'s data? A 0.70 °C/m B. 9.3 /m C. 4.2 °C/m D. 0.11 °C/m

Solution

The freezing points can be obtained from the curve by extrapolation. Extrapolate the portion of the temperature vs time plot where the temperature begins to stay constant. A rough estimation tells us that this should be -3.0C for pure water and -10.0C for the solution of sugar in water.

The depression in freezing point is T = T_pure – T_mix = (-3.0C) – (-10.0C) = 7.0C.

Molar mass of sugar, C₁₂H₂₂O₁₁ = (12*12.01 + 22*1.008 + 11*15.9994) g/mol = 342.2894 g/mol.

Moles of sugar = (6.0046 g)/(342.2894 g/mol) = 0.01754 mol.

Assume the density of water to be 1.00 g/mL; therefore, mass of 25.00 mL of water = (25.00 mL)*(1.00 g/mL) = 25.00 g = (25.00 g)*(1 kg/1000 g) = 0.025 kg.

Molality of sugar solution = (mol of sugar)/(kg of water) = (0.01754 mol)/(0.025 kg) = 0.7016m.

Use Raoult’s law for depression of freezing point.

T = k_f*(molality of solution)

===> 7.0 C = k_f*(0.7016m)

===> k_f = (7.0C)/(0.7016m) = 9.977 C/m

The closest match is 9.3C/m and hence we select this one.

Ans: (B).