Chemistry HW help

Question

You have an aqueous solution of NaCI that has a freezing point of - 5.25 °C. What mass (in grams) of NaCI must be added to 1000 g of this solution to lower the freezing point to -12.00 C•? (Assume a van't Hoff factor of 1.9 for NaCI). (Kf for water is 1.86 C•kg/mol).

Answer 1

∆T = imK

∆T = change in freezing point

i = van't Hoff factor

m = molality = mols NaCl / kg water

K = freezing constant = 186º/m

We need to start with how we got the solution with freezing point of -5.25º since normal freezing point is 0º.

∆T = 5.25º

i = 1.9

m = ? = mols NaCl / kg water

K = 1.86º/m

5.25 = (1.9)(m)(1.86)

m = 1.486 mols NaCl / kg water

1.486 mols NaCl / kg water x 58.44 g / mole = 86.84 g NaCl / kg water

Since this solution has a mass of 1000 g (1 kg), the mass of water must be less than 1 kg. What is the mass of water?

g NaCl + g water = 1000 g

86.84 g NaCl + 1000 g water = 1087 g

79.89 g NaCl + 919.96 g water = 1000 g

Thus, this solution contains 79.89 g NaCl and 919.96 g water

Moving on to the 2nd solution. How much NaCl needs to be added to lower freezing point from -5.25º to -12º?

∆T = imK

∆T = 6.75º

i = 1.9

m = ?

K = 1.86

6.75 = (1.9)(m)(1.86)

m = 1.91 mols NaCl / kg water

The mass of water (from above calculations) = 0.9196 kg

1.91 mols NaCl / kg water x 0.9196 kg water = 1.756 mols NaCl

1.756 mols NaCl x 58.44 g / mol = 102.6 g NaCl = 103 g NaCl need to be added

To check this, we can find the change in freezing point from adding 79.89 g of NaCl and 103 g NaCl to 0.9196 kg of water, and see if we get 12º

∆T = imK

∆T = ?

i = 1.9

m = 79.89 g NaCl + 103 g NaCl = 183 g x 1 mol / 58.44 g = 3.13 mols / 0.9196 kg = 3.41 m

K = 1.86

∆T = (1.9)(3.41)(1.86)

∆T = 12.0 Check.