Question

A solution containing 82.6 g KCl in 205 g H2O at 50 ∘C is cooled to 20 ∘C . Use the solubility data from the table below. Solubility (g/100gH2O) Substance 20∘C 50∘C KCl 34.0 42.6 NaNO3 88.0 114.0 C12H22O11 (sugar) 203.9 260.4 part A How many grams of KCl remain in solution at 20 ∘C Part B How many grams of solid KCl came out of solution after cooling

Answer 1

Ans. #A. Given, solubility of KCl in water at 20.0⁰C = 34.0 g KCl / 100 g water.

That is, 100.0 g water can solvate a maximum of 34.0 g KCl in it.

Now,

Mass of KCl solvated in 205 g water at 20.0⁰C = Solubility at 20.0⁰C x Mass of water

                                                            = (34.0 g KCl / 100 g water) x 205 g water

                                                            = 69.70 g KCl

Therefore, at 20.0⁰C, a total of 69.70 g KCl will remains solvated in 205 g water.

#B. Given, mass of KCl solvated in 205 g water at 50.0⁰C = 82.6 g

Mass of KCl solvated in 205 g water at 20.0⁰C = 69.7 g

Now,

Mass of KCl precipitated during solution the solution (82.6 g KCl/ 205 g water) to 20.0⁰C is equal to –

                        Mass of KCl in solution at 50.0⁰C – Maximum solvated KCl at 20.0⁰C

                        = 82.6 g – 69.7 g

                        = 12.9 g