# Chemistry = Freezing Point & Molality

• meganw
In summary, the problem involves finding the percent by mass of sugar in a mixture of sugar and table salt that causes a freezing point of -2.24 degrees Celsius when dissolved in 150g of water. The equation used is delta T = (i) (Kf solvent) (m) and the mass % is calculated by dividing the mass of one substance by the total mass of all substances. The attempt at a solution involves finding the molality of the salt and using it to calculate the amount of salt and sugar in the mixture. However, the calculated answer of 49.2% does not match the expected answer of 53.8%, leading to uncertainty about the accuracy of the solution.

## 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! =)

Sugar doesn't change the freezing point?

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.

## 1. What is the freezing point of a substance?

The freezing point of a substance is the temperature at which it changes from a liquid to a solid state. It is also known as the melting point, as it is the temperature at which the substance changes from a solid to a liquid state when heated.

## 2. How does molality affect the freezing point of a solution?

Molality is a measure of the concentration of a solution, expressed as the number of moles of solute per kilogram of solvent. The greater the molality of a solution, the lower its freezing point will be. This is because the presence of solute particles disrupts the crystal lattice structure of the solvent, making it more difficult for the solvent molecules to arrange themselves in a solid state.

## 3. What is the difference between molality and molarity?

Molality and molarity are both measures of solution concentration, but they differ in how they are calculated. Molality is calculated by dividing the number of moles of solute by the mass of the solvent in kilograms, while molarity is calculated by dividing the number of moles of solute by the volume of the solution in liters. In general, molality is a more accurate measure of concentration when dealing with solutions at different temperatures.

## 4. Can the freezing point of a substance be predicted?

Yes, the freezing point of a substance can be predicted using colligative properties, which are physical properties of a solution that depend on the number of solute particles present, rather than the type of solute. By knowing the molality of a solution and the freezing point depression constant, the freezing point of a solution can be calculated using the equation: ΔT = Kf * m, where ΔT is the change in freezing point, Kf is the freezing point depression constant, and m is the molality of the solution.

## 5. How does pressure affect the freezing point of a substance?

According to the phase diagram of a substance, an increase in pressure will cause the freezing point of a substance to decrease. This is because an increase in pressure favors the formation of a denser solid, which is achieved at a lower temperature. However, this effect is only significant at high pressures and is negligible in most cases.