Question

A 13.13 g mixture of sugar (C12H22O11) and table salt (NaCl) is dissolved in 253 g of water. The freezing point of the solution was measured as -2.41 °C. Calculate the mass percent of sugar in the mixture.

Solvent Formula Kf value* (°C/m) Normal freezing point (°C) Kb value (°C/m) Normal boiling point (°C)

water H2O 1.86 0.00 0.512 100.00

Answer 1

we have formula

dT = i x Kf x m

since we have two solutes , we do for two solutes seperately assuming eahc solute is 100 % and find Freezing points. Later we compare it real Frezzing point and find the fraction of each.

First glucose , moles of glucose = mass/ molar mass of glucsoe = 13.13/342.3 = 0.03836

molality = moles / solvent mass in kg = ( 0.03836 /0.253) = 0.151613

i = 1 for sucrose e since sugar is non dissociative.

now dT = 0 - Tf(solution) = 1 x 1.86 x 0.151613

Tf (solution) = -0.282 ......(1)

for NaCl , i = 2 since NaCl is dissociates to Na+ and Cl- ,

moles of NaCl = ( 13.13/58.44) = 0.224675

Molality = ( 0.224675 /0.253) = 0.888

Now 0 - Tf(solution) = 2 x 0.888 x 1.86

Tf(solution) = -3.3 .......(2)

Now we have Tf(solution ) actual = - 2.41 ............(3)

by (1) , (2) ( 3)

actual = fraction of sugar x FP of glucose solution + Fractionof NaCl x FP of NaCl solution

( let sugar fraction = X then NaCl fraction is 1-X)

-2.41 = ( X ) ( -0.282) + ( 1-X) ( -3.3)

2.41 = 0.282X + 3.3-3.3X

X = fractionof sugar = 0.295

Mass % of sugar = 100 x fraction =100 x 0.295 = 29.5 %