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.
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!!! =)