Suppose you wanted to know how many grams of KCl would be left if 350 mL of a 2.0 M KCl solution..

... solution were evaporated o dryness.

Suppose the molar mass of KCl is 74.55 grams per mole.

a) How would heating the solution affect the mass of KCl remaining? ... ( would it make it saturated???)
b) How many grams of KCl would remain?

This is the last chemistry problem on my homework and I can't figure it out :( Please help??

Problem A
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V = 350 mL = 350 * [1 Liter/1000 mL] = 0.350 L
C = 2.0 moles / L
n = C*V
n = 2 * 0.35 = 0.70 moles

1 mole KCl has a mass of 74.55
0.70 moles of KCl has a mass of x

x = 0.7 * 74.55
x = 52.2 grams.

Sorry,I thought you asked about the amount remaining after evaporation. Heating it might catch what little water remains, but it doesn't saturate it or change it chemically. The only way to get KCl to break down is to put it in an electrolytic cell, heat it enough to melt it and run electricity through it -- not an easy thing to do.

Problem B
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I knew the calculation would come in handy. 52.2 grams would remain.

a) The amount of KCl is constant, just that some is in solution and (as the solvent evaporates) some/all will deposit as solid.

b) Start with 1 litre of 1.0 M solution contains 1 mole of solute.
ie 1000 mL of 1.0M solution contains 1 mole of solute.
So
1000 mL of 2.0M solution contains 2 moles of solute
So
1mL of 2.0M solution contains 2/1000 moles of solute (= 0.002moles of solute)
So
350 mL of 2.0M solution contains (2/1000) x 350 moles of solute (0.7 moles of solute)
You know that 1 mole of KCl has the mass 74.55g
so 0.7 moles KCl has the mass 0.7 x 74.55g =52.185 (round to 52.19g)
All 52.19g of KCl will be deposited when the solution is evaporated to dryness.

a) The heating would not affect the mass of KCl in any way.

b)
(0.350 L) x (2.0 mol/L) x (74.55 g KCl/mol) = 52 g KCl