Question

c.In preparation for a different solubility determination, a student measured the mass of an empty beaker (25.62 g) on a balance. Then, KNO3 was added to the beaker and the mass recorded again as 35.18 g. Finally, water was added to the flask and the mass determined to be 50.03 g. The temperature of the mixture was increased until all of the solid had just dissolved. What was the solubility at this final temperature?

Answer 1

Mass of the empty beaker = 25.62g

Mass of the beaker filled with KNO3 = 35.18g

Mass of beaker filled with water and KNO3 = 50.03g

Therefore, Mass of KNO3 = 35.18 - 25.62 = 9.56 g

                   Mass of Water = 50.03 - 35.18 = 14.85g
                   volume of water = 14.85 ml (as density of water is 1g/cc)
                                              = 0.0148 L

Molecular mass of KNO3 = 39 + 14 + 48 = 101g

Moles of KNO3 = Given mass / molecular mass
                          = 9.56/101
                          =0.094 moles

Solubility is actually the concentration of KNO3 = moles / volume (in L)

                                                                            = 0.094 / 0.0148

                                                           Solubility = 6.351 mol/Litre