Answer:
The answer to your question is: letter B is correct
Explanation:
Calculate the Molarity of each solution an compare it to the solubility
Solubility
MW Cd(CN)2 = 252 g
252 g ---------------- 1 mol
1.7 g ------------------ x
x = (1.7 x1 ) / 252
x = 0.006 mol
Molarity = 0.006 / 0.1 = 0.067
a.97.0 g of Cd(CN)2 in 5.00 mL
252 g ---------------- 1 mol
97 g -------------------- x
x = (97 x 1) / 252
x = 0.38
Molarity = 0.38 / 0.005 =75 Saturated
b. 325.5 g of Cd(CN)2 in 20.5 L
252 g ---------------- 1 mol
325.5 g ---------------- x
x = (325.5 x 1) / 252
x = 1.29 mol
Molarity = 1.29 / 20.5 = 0.063 Unsaturated
c. No right answer.
d. 4.25 g of Cd(CN)2 in 250.0 mL
252 g ---------------- 1 mol
4.25 g --------------- x
x = (4.25 x 1) / 252
x = 0.017
Molarity = 0.017 / 0.25 = 0.067 Saturated
e. 5.750 g of Cd(CN)2 in 330.0 mL
252 g ---------------- 1 mol
5.75 g ------------------ x
x = (5.75 x 1) / 252 = 0.023
Molarity = 0.023 / 0.33 = 0.07 Saturated
