Problem #1: If you dilute 175 mL of a 1.6 M solution of LiCl to 1.0 L, determine the new concentration of the solution.
Solution:
M1V1 = M2V2
(1.6 mol/L) (175 mL) = (x) (1000 mL)
x = 0.28 M
Note that 1000 mL was used rather than 1.0 L. Remember to keep the volume units consistent.
Problem #2: You need to make 10.0 L of 1.2 M KNO3. What molarity would the potassium nitrate solution need to be if you were to use only 2.5 L of it?
Solution:
M1V1 = M2V2
(x) (2.5 L) = (1.2 mol/L) (10.0 L)
x = 4.8 M
Please note how I use the molarity unit, mol/L, in the calculation rather than the molarity symbol, M.
Problem #3: How many milliliters of 5.0 M copper(II) sulfate solution must be added to 160 mL of water to achieve a 0.30 M copper(II) sulfate solution?
Solution:
M1V1 = M2V2
(5.00 mol/L) (x) = (0.3 mol/L) (160 + x)
5x = 48 + 0.3x
4.7x = 48
x = 10. mL (to two sig figs)
The solution to this problem assumes that the volumes are additive. That's the '160 + x' that is V2.
Problem #4: What volume of 4.50 M HCl can be made by mixing 5.65 M HCl with 250.0 mL of 3.55 M HCl?
Solution:
Here is the first way to solve this problem:
M1V1 + M2V2 = M3V3
(3.55) (0.250) + (5.65) (x) = (4.50) (0.250 + x)
Where x is volume of 5.65 M HCl that is added
(0.250 + x) is total resultant volume
0.8875 + 5.65x = 1.125 + 4.50 x
1.15x = 0.2375
x= 0.2065 L
Total amount of 4.50 M HCl is then (0.250 + 0.2065) = 0.4565 L
Total amount = 456.5 mL
Here is the second way to solve this problem:
Since the amount of 5.65 M added is not asked for, there is no need to solve for it.
M1V1 + M2V2 = M3V3
(3.55) (250) + (5.65) (x − 250) = (4.50) (x)
That way, x is the answer you want, the final volume of the solution, rather than x being the amount of 5.65 M solution that is added.
Problem #5: A 40.0 mL volume of 1.80 M Fe(NO3)3 is mixed with 21.5 mL of 0.808M Fe(NO3)3 solution. Calculate the molar concentration of the final solution.
