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18 votes
What is the freezing point, in °C,
of a 0.520 m aqueous
solution of NaCl?

What is the freezing point, in °C, of a 0.520 m aqueous solution of NaCl?-example-1
User Cole Maclean
by
2.8k points

2 Answers

22 votes
22 votes

Answer: The freezing point of a solution is the temperature at which the solution becomes a solid. The freezing point of a solution is lower than the freezing point of the pure solvent because the solute particles interfere with the movement of the solvent molecules, which slows down the freezing process.

To determine the freezing point of a solution, we can use the freezing point depression equation:

ΔTf = Kf x molality

where ΔTf is the change in freezing point, Kf is the freezing point depression constant for the solvent, and molality is the concentration of the solute in the solution expressed in moles of solute per kilogram of solvent.

To find the freezing point of a 0.520 m aqueous solution of NaCl (sodium chloride), we need to know the freezing point depression constant for water, which is 1.86 °C/m. We can then use the equation above to calculate the change in freezing point:

ΔTf = 1.86 °C/m x 0.520 m = 0.97 °C

To find the freezing point of the solution, we need to subtract the change in freezing point from the freezing point of the pure solvent. The freezing point of pure water is 0 °C, so the freezing point of the 0.520 m aqueous solution of NaCl is:

0 °C - 0.97 °C = -0.97 °C

User Judoole
by
3.3k points
29 votes
29 votes

Answer:

-1.934

Step-by-step explanation:

0 - {(0.520) x (1.86) x (2)}

User Nickknack
by
2.7k points