Worksheet - Dilution - Problems 1 - 10


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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:

M1aV1a + M1bV1b = M2V2

(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.

M1aV1a + M1bV1b = M2V2

(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.

Solution:

Let's use a slightly different way to write the subscripts:

M1V1 + M2V2 = M3V3

There is no standard way to write the subscripts in problems of this type.

Substituting:

(1.80) (40.0) + (0.808) (21.5) = (M3) (40.0 + 21.5)

M3 = 1.45 M


Problem #6: To 2.00 L of 0.445 M HCl, you add 3.88 L of a second HCl solution of an unknown concentration. The resulting solution is 0.974 M. Assuming the volumes are additive, calculate the molarity of the second HCl solution.

Solution #1:

M1V1 + M2V2 = M3V3

(0.445) (2.00) + (x) (3.88) = (0.974) (2.00 + 3.88)

x = 0.125 M

Solution #2:

1) Calculate moles HCl in 0.445 M solution:

(0.445 mol/L) (2.00 L) = 0.890 moles

2) Set up expression for moles of HCl in second solution:

(x) (3.88 L) = moles HCl in unknown solution

3) Calculate moles of HCl in final solution:

(0.974 mol/L) (5.88 L) = 5.73 moles

4) Moles of HCl in two mixed solutions = moles of HCl in final solution:

0.890 moles + [(x) (3.88 L)] = 5.73 moles

x = 1.25 M (to three sig figs)


Problem #7: To what volume should you dilute 133 mL of an 7.90 M CuCl2 solution so that 51.5 mL of the diluted solution contains 4.49 g CuCl2?

Solution:

1) Find moles:

(4.49g CuCl2) (1 mole CuCl2 / 134.45 grams) = 0.033395 moles CuCl2

2) Find the molarity of the 51.5 mL of the diluted solution that contains 4.49g CuCl2:

(0.033395 moles CuCl2) / (0.0515 liters) = 0.648 M

3) Use the dilution formula:

M1V1 = M2V2

(7.90 M) (133 mL) = (0.648 M) (V2)

V2 = 1620 mL

You should dilute the 133 mL of an 7.90 M CuCl2 solution to 1620 mL.


Problem #8: If volumes are additive and 95.0 mL of 0.55 M KBr is mixed with 165.0 of a BaBr2 solution to give a new solution in which [Br¯] is 0.65 M, what is the concentration of the BaBr2 used to make the new solution?

Solution:

moles of Br¯ from KBr: (0.55 mol/L) (0.095 L) = 0.05225 mol

moles of Br¯ in final solution: (0.65 mol/L) (0.260 L) = 0.169 mol

moles Br¯ provided by the BaBr2 solution: 0.169 - 0.05225 = 0.11675 mol

BaBr2 provides two Br¯ per formula unit so (0.11675 divided by 2) moles of BaBr2 are required for 0.11675 moles of Br¯ in the solution.

molarity of BaBr2 solution: 0.058375 mol / 0.165 L = 0.35 M


Problem #9: 1.00 L of a solution is prepared by dissolving 125.6 g of NaF in it. If I took 180 mL of that solution and diluted it to 500 mL, determine the molarity of the resulting solution.

Solution:

1) Calculate moles of NaF:

125.6 g / 41.9 g/mol = 3.00 mol

2) Calculate moles in 180 mL of resulting solution:

3.00 mol in 1000 mL so 3 x (180/1000) = 0.54 mol in 180 mL

3) Calculate molarity of diluted solution:

0.54 mol / 0.50 L = 1.08 mol/L = 1.08 M

Problem #10: What is the molar concentration of chloride ions in a solution prepared by mixing 100.0 mL of 2.0 M KCl with 50.0 mL of a 1.50 M CaCl2 solution?

(Warning: there's a complication in the solution. It has to do with the CaCl2.)

Solution #1:

1) Get total moles of chloride:

KCl ⇒ (2.00 mol/L) (0.100 L) = 0.200 mol of chloride ion

CaCl2 ⇒ (1.50 mol/L) (0.0500 L) (2 ions / 1 formula unit) = 0.150 mol of chloride ion.

The '2 ions / 1 formula unit' is the problem child. The solution is 1.50 M in calcium chloride, but 3.00 M in just chloride ion.

total moles = 0.200 mol + 0.150 mol = 0.350 mol

2) Get chloride molarity:

0.350 mol / 0.150 L = 2.33 M

Solution #2:

Suppose you really wanted to use this equation:

M1V1 + M2V2 = M3V3

Set it up like this:

(2.00 mol/L) (0.100 L) + (3.00 mol/L) (0.0500 L) = (M3) (0.150 L)

Note that the CaCl2 molarity is 3.00 because that is the molarity of the solution from the point-of-view of the chloride ion.


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