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Given the volumes of a solvent and solute and their density, what temperature do they freeze at? It would really help if someone helped explain how to do this, I’m kind of stuck thanks

Given the volumes of a solvent and solute and their density, what temperature do they-example-1

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Answer:

Step-by-step explanation:

To determine the freezing temperature of a solvent and solute mixture, you will need to use the freezing point depression equation. This equation allows you to calculate the change in freezing point of a solvent caused by the addition of a solute.

The freezing point depression equation is:

ΔTf = Kf x molality

where:

ΔTf is the change in freezing point of the solvent in degrees Celsius

Kf is the freezing point depression constant for the solvent, which is a specific value for each solvent and is usually provided in a table or chart

molality is the concentration of the solute in the solvent, expressed in mol/kg

To use this equation, you will need to know the volume and density of the solvent and solute, as well as the molecular weight of the solute. You can then use these values to calculate the molality of the solute in the solvent.

For example, let's say you have a mixture of 100 mL of water (solvent) and 10 g of salt (solute). The density of water is 1 g/mL and the molecular weight of salt is 58.44 g/mol.

To calculate the molality of the salt in the water, you can use the following steps:

Calculate the mass of the solvent: 100 mL x 1 g/mL = 100 g

Calculate the molality of the solute: 10 g / 58.44 g/mol = 0.171 mol

Calculate the molality of the solvent: 0.171 mol / 100 g = 0.00171 mol/g

Once you have calculated the molality of the solute in the solvent, you can use the freezing point depression equation to calculate the change in freezing point of the solvent caused by the addition of the solute.

For water, the freezing point depression constant (Kf) is 1.86 °C/m. Therefore, the change in freezing point caused by the addition of the salt to the water would be:

ΔTf = 1.86 °C/m x 0.00171 m = 0.00316 °C

This means that the freezing temperature of the solvent and solute mixture would be 0.00316 °C lower than the freezing temperature of pure water, which is 0 °C. Therefore, the freezing temperature of the mixture would be -0.00316 °C.

I hope this helps! Let me know if you have any questions or need further clarification.

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