You will use molarity and molality to solve problem three different ways
1. determine molarity or molality
2. find out how many much solution is needed to get a number of moles of solute
3. calculate how much solute is needed to make a specified volume of solution
1. What is the molarity of a solution made by dissolving 3.5 grams of NaCl in 250
ml of water?
You need to find the number of moles and divide by the number of liters.
Convert grams to moles NaCl then ml of water the liters
3.5 grams NaCl
1 mole
58.5 g NaCl
1000ml
250ml
1L
answer: 0.239 or with sig figs 0.24 M (M for molarity, moles per liter)
Molality works the same, just use kg of solvent instead of liters, m for molality.
2
How many milliliters of a 1.5 M solution is needed to get .075 moles of solute?
Moles = molarity times volume(in liters)
1.5 M x V = .075
V= .075 / 1.5 = .050 liters = 50 ml
2. How much NaCl is needed to make 350 ml of 1.2 M solution?
Moles = molarity times volume( in liters)
= 1.2 x 0.350 = 0.420 moles
now change moles to grams
0.420 mol NaCl
58.5 grams NaCl
1 mole NaCl
=
24.6 grams NaCl dissolve in enough
water to make 350 ml solution
Colligative Properties
The molality of a solution effects the freezing and boiling points of a solvent.
Boiling point is increased and freezing point is depressed(decreased)
Tb=mKb
Tf=mKf
What is the freezing point depression of a 1.50 m solution, if Kf is 1.86deg /m?
Tf=1.50m x 1.86 oC/m =2.8 degrees C
What is the new freezing point if the solvent is water?
SUBTRACT the depression from the original freezing point.
Tf = 0-2.8 = -2.8 degrees C
Boiling point elevation works the same but the boiling temp is ADDED to the normal
boiling point. Also the boiling point constant will not be the same as the freezing point
constant.
What is the boiling point elevation of a 5.50 m solution, if Kb is 0.52 deg /m?
Tf=5.50m x 0.52 oC/m =2.6 degrees C
What is the new boiling point if the solvent is water?
ADD the elevation to the original boiling point.
Tf = 100+2.6 = 100.26 degrees C