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**

1) A solution was prepared by dissolving 489
milligrams of potassium sulfate(mw=174.24 g/mol) in 537mL of water.
Calculate the following:
a) [K+]
b) % (w/v) K2SO4
c) pK+
d) pSO4^-2

1) A solution made by dissolving 0.658 moles of non
electrolyte solution in 855 grams of benzene. Calculate the
freezing point of the solution.
2) write the equilibrium constant expression for the reaction:
A(s)+ 3B(l)⇌2C(aq)+D (aq) in terms of [A], [B], [C] and [D] as
needed.
3) The equilibrium constant for the chemical equation:
N2(g)+3H2(g)⇌2NH3(g) is Kp=0.0451 at 235 degrees Celsius. Calculate
the value of the Kc for the reaction at 235 degrees Celsius.

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