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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

1. [pic]

2. 75.0 mL

3. [pic]

4. 15,000 L or 1.5 x 107 mL

5.

[pic]

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 [pic]

Step 3: [pic]

Could this be done as a single DA? Yes but it would be messy and hard to follow.

How do you calculate molarity after dilution? Where M2 is the final molarity (moles/L) M1 is the initial molarity (moles/L) V1 is the initial volume (L) V2 is the final volume (L)