The solubility of ammonium permanganate is 7.90 g/100 mL of water.29.0 grams of ammonium permanganate are placed in a container and 220 mL of water is added.The liquid is allowed to stand in contact with the solid until equilibrium is achieved and then the liquid is poured off.What is the weight of solid left?
A)17.4 g
B)79.0 g
C)7.90 g
D)11.6 g
E)29.0 g

Question

Quizplus · Accepted Answer

First, we need to determine how much of the ammonium permanganate will actually dissolve in the 220 mL of water.

From the solubility data given, we know that 100 mL of water can dissolve 7.90 g of ammonium permanganate. Therefore, 220 mL of water can dissolve:

(7.90 g/100 mL) x (220 mL) = 17.38 g of ammonium permanganate

Since we added 29.0 g of ammonium permanganate to the water, we know that not all of it will dissolve. The amount left behind will be the initial amount added (29.0 g) minus the amount that actually dissolves (17.38 g).

29.0 g - 17.38 g = 11.62 g

Therefore, the weight of solid left behind is 11.6 g, which corresponds to answer choice D.

The Solubility of Ammonium Permanganate Is 7