Question

4) At 341K, this reaction has a Kc value of 0.0164.     2X (g)+ 3Y (g)  ⇌2Z (g)   Calculate Kp at 341K.
5) a solution is made by dissolving 0.658 moles of nonelectrolyte solute in 855 grams of benzene. Calculate the freezing point of the solution.
6) a solution was prepared by dissolving 489 milligrams of potassium sulfate(K2SO4,MW=174.24 g/mol) in 537 milliliters of water .calculate the following: a) [K+]   b)pK+   c) pSO4^2-

Answer 1

4)

Kc = 0.0164

delata n = 2 - 5 = -3

Kp = Kc (RT)^Dn

     = 0.0164 x (0.0821 x 341)^-3

Kp = 7.47 x 10^-7

5)

molality = 0.658 / 0.855

             = 0.769

delta Tf = Kf x m

To - Tf = 5.12 x 0.769

5.5 - Tf = 3.94

Tf = 1.56 oC

freezing point of solution = 1.56 oC

6)

moles of K2SO4 = 489 x 10^-3 / 174.24 = 2.81 x 10^-3

Molarity = 2.81 x 10^-3 / 0.537

              = 0.0052 M

[SO42-] = 0.0052 M

[K+] = 2 x 0.0052 = 0.010 M

pK+ = - log [K+] = 2

pSO42- = - log [SO42-] = 2.28