# Chemistry = Freezing Point & Molality

## Homework Statement

You have a 10.4g mixture of sugar (C12H22O11) and table salt (NaCl). When this mixture is dissolved in 150g of water the freezing point is found to be –2.24 degrees centigrade. Calculate the percent by mass of sugar in the original mixture.

## Homework Equations

delta T = (i) (Kf solvent)(m)
Mass % = mass of substance 1 / total mass of all substances

## The Attempt at a Solution

First I tried to find the molality of the salt
delta T = 2.24 = 2 (1.86) (m)
m=.602151= mols salt/kg water

so .602151m = X mols salt/.15 kg water

X mols of salt = .090323
grams of salt = 5.27872 grams of salt.

So....grams of sugar in the mixture would be 5.12128 grams sugar.

The percent mass by sugar is then 5.12128/10.40....but that's 49.2% and our teacher (the exam key) said that the answer was 53.8%.

Does anyone know what I am doing wrong? Thanks so much!!! =)

## Answers and Replies

Borek
Mentor
Sugar doesn't change the freezing point?