Name _______KEY___________________________________ Date _________ Period ______
Molarity and Dilution
1.
What is the molarity of a 0.77 L solution with 5.5 moles of sodium chloride?
M=
2.
mol 5.5 mol
=
= 7.1 M
V
0.77 L
What is the molarity if 16.5 g of copper (II) sulfate are dissolved in 2.25 L of solution?
16.5 g CuSO4 x
M=
3.
1 mol CuSO4
= 0.103 mol CuSO4
159.62 g CuSO4
mol 0.103 mol
=
= 0.0459 M
V
2.25 L
How many moles of Na2CO3 are needed to make 250. mL of a 2.50 M solution?
mol = MV = (2.50 M)(0.250 L) = 0.625 mol
4.
How many grams of aluminum fluoride are needed to make 5.15 L of a 1.45 M solution?
mol = MV = (1.45 M)(5.15 L) = 7.47 mol
7.47 mol AlF3 x
5.
What is the volume of a 6.75 M solution containing 23.7 g of HCl?
23.7 g HCl x
V=
6.
83.98 g AlF3
= 627 g AlF3
1 mol AlF3
1 mol HCl
= 0.650 mol HCl
36.46 g HCl
mol 0.650 mol
=
= 0.0963 L
M
6.75 M
How many milliliters of 4.00 M KI are needed to prepare 250 mL of 0.760 M KI?
M1V1 = M2V2 ; V2 =
7.
I have 345 mL of a 1.50 M CoO solution. If I boil away water until the volume of the
solution is 250. mL, what will the molarity of the solution be?
M1V1 = M2V2 ; M2 =
8.
M1V1 (1.50 M)(0.345 L)
=
= 2.07 M
V2
0.250 L
What is the concentration when you add 563 mL of water to 341 mL of a 0.250 M
Ba(OH)2 solution?
M1V1 = M2V2 ; M2 =
9.
M1V1 (0.760 M)(0.250 L)
=
= 0.048 L = 48 mL
M2
4.00 M
M1V1 (0.250 M)(0.341 L)
=
= 0.0943 M
V2
(0.341 L + 0.563 L)
If 0.750 L of 0.450 M sugar solution is left uncovered on a windowsill, and 193 mL of
the solvent evaporates, what is the new concentration of the solution?
M1V1 = M2V2 ; M2 =
M1V1 (0.450 M)(0.750 L)
=
= 0.606 M
V2
(0.750 L - 0.193 L)