#### Answer

see solution

#### Work Step by Step

(a) 30.0 g $KClO_3$ in 85.0 g of water at 35 °C
30 g in 85 g of water
= (85 X 30)/100
= 25.5 g in 100 g of water
The solubility of $KClO_3$ at 35 °C = 12 g / 100 g of water.
So, the given solution is a supersaturated solution and cannot dissolve the solute.
(b) 65.0 g $NaNO_3$ in 125 g of water at 15 °C
65.0 g $NaNO_3$ in 125 g of water
= (65 X 125)/100
= 81.25 g in 100 g of water
The solubility of $NaNO_3$ in 100 g of water at 15 °C = 84 g
So, the given solution is unsaturated and can completely dissolve the solute molecules.
(c) 32.0 g $KCl$ in 70.0 g of water at 82 °C
32.0 g $KCl$ in 70.0 g of water
= (32 X 70) / 100
= 22.4 g in 100 g of water
Solubility of $KCl$ in 100 g of water at 82 °C = 51 g.
So, the given solution is unsaturated and can completely dissolve the solute molecules.