The following questions all refer to an aqueous 8.99 M sodium chlorite stock sol
ID: 911309 • Letter: T
Question
The following questions all refer to an aqueous 8.99 M sodium chlorite stock solution.
1. If 14.5 mL of the stock solution are diluted to 538 mL, what is the molarity (in mol/L) of the new solution?
2. If 16.8 mL of the stock solution are diluted to obtain a 0.237 M sodium chlorite solution, what is the volume (in mL) of the new solution?
3. Determine the volume (in mL) of the original stock solution that must be used to prepare 715 mL of a 0.320 M sodium chlorite solution.
4. Determine the volume of water (in mL) that was added to 11.5 mL of the stock solution to prepare a 0.626 M sodium chlorite solution.
Explanation / Answer
1. Molarity of the stock solution, M1 = 8.99 M
Volume of the stock solution diluted, V1 = 14.5 mL
Molarity of the diluted solution, M2 = ?
Volume of the diluted solution, V2 = 538 mL
Using the dilution law, M1V1 = M2V2
(8.99 M) (14.5mL) = M2 (538mL)
M2 = 0.242 M
2. Molarity of the stock solution, M1 = 8.99 M
Volume of the stock solution diluted, V1 = 16.8 mL
Molarity of the diluted solution, M2 = 0.237 M
Volume of the diluted solution, V2 =?
Using the dilution law, M1V1 = M2V2
(8.99 M) (16.8mL) = 0.237 M(V2)
V2 = 637.27 mL
3.
Molarity of the stock solution, M1 = 8.99 M
Volume of the stock solution diluted, V1 = ?
Molarity of the diluted solution, M2 = 0.320 M
Volume of the diluted solution, V2 = 715 mL
Using the dilution law, M1V1 = M2V2
(8.99 M) (V1) = 0.320 M(715 mL)
V1 = 25.45 mL
4.
Molarity of the stock solution, M1 = 8.99 M
Volume of the stock solution diluted, V1 = 11.5 mL
Molarity of the diluted solution, M2 = 0.626 M
Volume of the diluted solution, V2 =?
Using the dilution law, M1V1 = M2V2
(8.99 M) (11.5mL) = 0.626 M(V2)
V2 = 165.15 mL
Volume of water added = 165.15 mL - 11.5 mL = 153.65 mL
153.65 mL water is to be added.
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