Chemistry = Freezing Point & Molality

[meganw]

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! =)

 

[Borek]

Sugar doesn't change the freezing point?

 

[Blueshift5]

Based on your calculations, it seems that you have made a mistake in converting from moles to grams. When converting from moles to grams, you need to multiply by the molar mass of the substance. In this case, the molar mass of salt (NaCl) is 58.44 g/mol and the molar mass of sugar (C12H22O11) is 342.3 g/mol. Therefore, the correct calculation for the grams of salt in the mixture is (0.090323 mol NaCl)(58.44 g/mol) = 5.27872 grams of salt. Similarly, the grams of sugar in the mixture would be (0.090323 mol C12H22O11)(342.3 g/mol) = 30.928 grams of sugar.

Using these values, the percent mass of sugar in the original mixture would be (30.928 g sugar)/(10.4 g mixture) x 100% = 297.38%. However, this is not a valid answer as it is above 100%. This is likely due to a mistake in your conversion from molality to moles. The correct calculation would be (0.602151 mol/kg water)(0.15 kg water) = 0.090323 moles of salt.

Using this value, the correct percent mass of sugar in the original mixture would be (30.928 g sugar)/(36.206 g total mixture) x 100% = 85.5%. This is much closer to the answer given by your teacher (53.8%), but there may be other factors at play such as rounding errors or experimental error. Overall, it is important to double check your conversions and calculations to ensure accuracy in your results.