Question

A 15.86 g mixture of sugar (C12H22O11) and table salt (NaCl) is dissolved in 145 g of water. The freezing point of the solution was measured as -4.45 °C. Calculate the mass percent of sugar in the mixture.

Answer 1

let n1 be no. of mol of sugar
and n2 be no. of mol of NaCl

i for sugar is 1
i for NaCl is 2
Kf for water = 1.86 oC/m
use,
(delta Tf) = Kf*{(n1 + 2*n2)/(mass of water in kg)}
0 - (-4.45) = 1.86*{(n1+2*n2)/0.145}
4.45 = 1.86*{(n1+2*n2)/0.145}
0.645 = 1.86*(n1+2*n2)
n1+2*n2 = 0.347

let mass of Sugar be x
then, mass of NaCl = 15.86-x

molar mass of Sugar(m1) = 342 g/mol
molar mass of NaCl(m2) = 58.5 g/mol

use,
(mass of sugar)/m1 + 2*(mass of salt)/m2 = 0.347
x/342 + 2*(15.86-x)/58.5 = 0.347
58.5*x + 10848.2 - 684*x = 6942.4
-625.5*x = -3905.8
x = 6.2 g

so,
mass of sugar = x
= 6.2 g

mass % of sugar = {(mass of sugar)/(mass of mixture)}*100
= (6.2/15.86)*100
= 39.1 %

Answer: 39.1%