Dilution Worksheet (Section 12.
3)
Concentration:
Dilution: To dilute a solution means to add more solvent without the addition of more solute. Of course, the
resulting solution is thoroughly mixed so as to ensure that all parts of the solution are identical. The fact that the
solute amount stays constant allows us to develop calculation techniques so that:
moles before dilution = moles after dilution
From the definition of molarity, we know that the moles of solute equals the molarity times the volume. So we can
substitute MV (molarity times volume) into the above equation, like this:
M1V1 = M2V2
where the left side of the equation is before dilution and the right side after dilution.
1. What is the concentration (molarity) of a solution of NaCl if 40. mL of a 2.5 M NaCl solution is diluted to a total
volume of 500. mL?
2. A stock solution of 1.00 M NaCl is available. How many mL are needed to make 100.0 mL of 0.750 M
3. What volume of 0.250 M KCl is needed to make 100.0 mL of 0.100 M solution?
4. To properly disposed of acid its concentration must be less than 1.00 x 10 -5 M. How much water must be added to
25 mL of 6.0 M HCl before it can be disposed of safely? (This is why we don’t pour it away).
5. 500. mL of a 6.00 M stock solution of NaCl is added to 2.00 L of water. How much of the solution must you pour
away and replace with water to get exactly 2.00 L of 1.00 M NaCl?
6. How much solvent must be added to 200. mL 1.50 M NaNO3 to make a solution with a concentration of 0.800 M
NaNO3 ?
7. Suppose you have just received a shipment of sodium carbonate, Na2CO3 . You weigh out 50.00 g of
the material, dissolve it in water, and dilute the solution to 1.000 L. You remove 10.00 mL from the solution
and dilute it to 50.00 mL. By measuring the amount of a second substance that reacts with Na2CO3 , you
determine that the concentration of sodium carbonate in the diluted solution is 0.0890 M. Calculate the percentage of
Na2CO3in the original batch of material.
Dilution Worksheet (Section 12.3)
Concentration:
Dilution: To dilute a solution means to add more solvent without the addition of more solute. Of course, the
resulting solution is thoroughly mixed so as to ensure that all parts of the solution are identical. The fact that the
solute amount stays constant allows us to develop calculation techniques so that:
moles before dilution = moles after dilution
From the definition of molarity, we know that the moles of solute equals the molarity times the volume. So we can
substitute MV (molarity times volume) into the above equation, like this:
M1V1 = M2V2
where the left side of the equation is before dilution and the right side after dilution.
1. What is the concentration (molarity) of a solution of NaCl if 40. mL of a 2.5 M NaCl solution is diluted to a total
volume of 500. mL?
2. A stock solution of 1.00 M NaCl is available. How many mL are needed to make 100.0 mL of 0.750 M
3. What volume of 0.250 M KCl is needed to make 100.0 mL of 0.100 M solution?
4. To properly disposed of acid its concentration must be less than 1.00 x 10 -5 M. How much water must be added to
25 mL of 6.0 M HCl before it can be disposed of safely? (This is why we don’t pour it away).
5. 500. mL of a 6.00 M stock solution of NaCl is added to 2.00 L of water. How much of the solution must you pour
away and replace with water to get exactly 2.00 L of 1.00 M NaCl?
6. How much solvent must be added to 200. mL 1.50 M NaNO3 to make a solution with a concentration of 0.800 M
NaNO3 ?
7. Suppose you have just received a shipment of sodium carbonate, Na2CO3. You weigh out 50.00 g of
the material, dissolve it in water, and dilute the solution to 1.000 L. You remove 10.00 mL from the solution
and dilute it to 50.00 mL. By measuring the amount of a second substance that reacts with Na2CO3, you determine that
the concentration of sodium carbonate in the diluted solution is 0.0890 M. Calculate the percentage of Na2CO3in the
original batch of material.
ANSWER KEY - DILUTION WS
M1V1 2.5 M x 40. mL
1. M2 0.20 M NaCl
V2 500. mL
2. 75.0 mL
M 2 V2 0.100 M x 100.0 mL
3. V1 40.0 mL KCl
M1 0.250 M
4. 15,000 L or 1.5 x 107 mL
5.
M1V1 6.00 M x 500.0 mL
M2 1.20 M NaCl
V2 2.50 L
now, you need to dilute again
M 2V2 1.00 M x 2.00 L
V1 1.67 L, so you need
M1 1.20 M
to pour out 2.50 -1.67 0.83 L (or 0.833 L)
6. final volume
is 375 mL, so you need to add 175 mL
7. To solve this, work backwards from the final dilution forward:
Step 1: #moles in final dilution = 0.0890M x 0.0500L = 0.00445
moles Na2CO3
Step 2: # mole in second dilution
0.00445 moles Na2CO 3
x 1.00L 0.445 moles Na2CO 3
0.01000 L
105.99g
0.445 moles Na 2CO 3 x
Step 3: 1 moles Na 2CO 3
x100% 94.3%
50.00g
Could this be done as a single DA? Yes but it would be messy and hard
to follow.
