Question

5. The solubility of KCl (in grams KCl/100 g of water) is 31.0 g at 10.0 °C, 34.0 g at 20.0 °C, 42.6 g at

50.0 °C, and 56.7 at 100 °C.

a) You want to make a saturated solution of KCl at 50.0 °C using 2.00 L of water. How many grams of KCl are necessary to make this solution? You may assume that the density of water is 1.00 g/mL.

b) You then take your saturated solution of KCl and 2.00 L of water (mixed) and place it in a refrigerator set at 10 °C. Yow much KCl (in grams) will have precipitated out of the solution once the solution has cooled down to 10 °C? You may assume that a supersaturated solution does not form.

Answer 1

5. a. Solubility of KCl at 50.0°C=42.6 g per 100 g water

We know that 1 L=1000 mL

So 2 L=2x1000 mL=2000 mL

Mass of 2000 mL water=Volume x density

=2000 mL x 1.00 g/mL

=2000 g

So at 50.0°C

100 g water can dissolve a maximum of 42.6 g KCl

So 1 g water can dissolve a maximum of (42.6/100) g KCl

And 2000 g water can dissolve a maximum of (42.6/100)x2000 g=852 g KCl

So 852 g KCl is required to make a saturated solution of KCl at 50.0°C in 2 L water.

b. Given at 10.0 °C, solubility of KCl =31.0 g per 100 g water

i.e. at 10.0°C

100 g water can dissolve a maximum of 31.0 g KCl

1 g water can dissolve a maximum of (31.0/100) g KCl

2000 g water can dissolve a maximum of (31.0/100)x2000 g= 620 g KCl

So out of the 852 g KCl that dissolved in 2000 g water at 50.0°C, only 620 g remained dissolved and the remaining (852-620)g=232 g KCl precipitated out of the solution once the saturated solution made at 50.0°C was cooled to 10.0°C.