1. A solution is made by dissolving 15.5 grams of glucose (C6H12O6) in 245 grams of water. What is the freezing-point depression of the solvent if the freezing point constant is -1.86 °C/m? Show all of the work needed to solve this problem. (5 points)

2.A solution is made by dissolving 2.5 moles of sodium chloride (NaCl) in 198 grams of water. If the molal boiling point constant for water (Kb) is 0.51 °C/m, what would be the boiling point of this solution? Show all of the work needed to solve this problem. (5 points)

I don't understand how to solve these.

Question

1. A solution is made by dissolving 15.5 grams of glucose (C6H12O6) in 245 grams of water. What is the freezing-point depression of the solvent if the freezing point constant is -1.86 °C/m? Show all of the work needed to solve this problem. (5 points)

2.A solution is made by dissolving 2.5 moles of sodium chloride (NaCl) in 198 grams of water. If the molal boiling point constant for water (Kb) is 0.51 °C/m, what would be the boiling point of this solution? Show all of the work needed to solve this problem. (5 points)



I don't understand how to solve these.

sweetburger · Answer

Molality(Freezing Point Constant) = Delta T
sooooo
15.5/180=.086/.245=.345m(-1.86C/m)= .65 so the normal freezing point of water is 0C and the solution depresses it by .65C so the new freezing point would be -.65C

sweetburger · Answer

for question 2 follow the same process as number one but consider Van hofts constant which would be 2 and the equation you would use is $$\Delta T=iKm$$

sweetburger · Answer

van hofts factor* not constant

anonymous · Answer

How do I use that equation?