Question

1. Calculate the vapor pressure of water over a solution of ethylene glycol, C2H6O2, at 22C which has a mole fraction of ethylene glycol of .288. The vapor pressure of pure water at 22C is 19.83mm-Hg.

2. What is the final potassium ion molarity in a solution made by mixing 300mL of 1M potassium nitrate and 700mL of 2M potassium phosphate (k3PO4)?

Answer 1

(1) According to Raoult's law,

Relative lowering of vapour pressure = mole fraction of solute

(p0 - p) / p0 = X_B

(19.83 - p) / 19.83 = 0.288

19.83 - p = 5.711

Therefore pressure of water over solution, p = 14.12 mm of Hg

(2) No. of moles of Potassium nitrate = 300 x 1/ 1000 = 0.3 mol

Then, number of moles of potassium ions produced from potassium nitrate = 0.3 mol

No. of moles of Potassium phosphate = 700 x 2 / 1000 = 1.4 mol

Then number of moles of potassium ions produced from potassium phosphate = 3 x 1.4 = 4.2 mol

Therefore total number of moles of potassium ions = 0.3 + 4.2 = 4.7 mol

Now, Molarity = Number of moles / volume of solution in L

M = 4.7 / 1 L

M = 4.7 M