Question

7. If 6.050 g of silver sulfate is mixed thoroughly with 300. mL of water. The silver sulfate that does not dissolve is filtered out. How many grams of Ag₂SO_4­ will be filtered out?

            A) 3.870                      B) 2.116                      C) 1.449               D) 4.629                      E) 5.072

Please show work! thank you! Answer: D

Answer 1

Ksp of Ag2SO4 = 1.2*10^-5

The salt dissolves as:

Ag2SO4 <----> 2 Ag+ + SO42-

   2s s

Ksp = [Ag+]^2[SO42-]

1.2*10^-5=(2s)^2*(s)

1.2*10^-5= 4(s)^3

s = 1.442*10^-2 M

Molar mass of Ag2SO4 = 2*MM(Ag) + 1*MM(S) + 4*MM(O)

= 2*107.9 + 1*32.07 + 4*16.0

= 311.87 g/mol

Molar mass of Ag2SO4= 311.87 g/mol

s = 1.442*10^-2 mol/L

To covert it to g/L, multiply it by molar mass

s = 1.442*10^-2 mol/L * 311.87 g/mol

s = 4.498 g/L

mass dissolved = s*volume

= 4.498 g/L * 0.300 L

= 1.349 g

mass that is filtered = initial mass - mass dissolved

= 6.050 g - 1.349 g

= 4.7 g

Answer: D