A student adds 50 g of KNO3 to 50 mL of water at 60 degrees C. The solution is cooled to 10 degrees C and then poured through a piece of 2 g filter paper in a funnel which is dried overnight. What is the predicted mass of the dried filter paper and solid?

So, you know that you're dissolving 50 g of potassium nitrate, #KNO_3#, a soluble salt, in 50 mL of water at a temperature of #60^@"C"#.

The first thing you need to determine is whether or not you can dissolve that much potassium nitrate in that much water. To do that, use potassium nitrate's solubility curve, which looks like this

Notice that, at #60^@"C"#, you can dissolve approximately 115 g of potassium nitrate in 100 g of water. To determine the mass of water you'd get at that temperature, use a water density calculator

http://antoine.frostburg.edu/chem/senese/javascript/water-density.html

At #60^@"C"#, water has a density of 0.9832 g/mL. This means that you have

#50cancel("mL") * "0.9832 g"/(1cancel("mL")) = "49.2 g water"#

So, at this temperature, you can dissolve

#49.2cancel("g water") * ("115 g "KNO_3)/(100cancel("g water")) = "56.6 g"# #KNO_3#

in that muc hwater. Since you have less potassium nitrate than that, your solution will be unsaturated.

When you cool the solution to #10^@"C"#, potassium nitrate's solubility decreases significantly. At this timperature, you can only dissolve approximately 25 g of potassium nitrate per 100 g of water.

This means that the excess potassium nitrate will precipitate out of the solution. At #10^@"C"#, the density of water is basically equal to 1 g/mL. This means that your solution can hold no more than

#50cancel("g water") * ("25 g "KNO_3)/(100cancel("g water")) = "12.5 g"# #KNO_3#

Therefore, the mass of the precipitate will be

#m_"precipitate" = 50 - 12.5 = "37.5 g"# #KNO_3#

Adding the mass of the dried paper filter will get you

#m_"total" = 2 + 37.5 = "39.5 g"#

Rounded to one sig fig, the answer will be

#m_"total" = color(green)("40 g")#