The following questions all refer to an aqueous 7.11 M potassium carbonate stock
ID: 910458 • Letter: T
Question
The following questions all refer to an aqueous 7.11 M potassium carbonate stock solution.
1. If 19.3 mL of the stock solution are diluted to 858 mL, what is the molarity (in mol/L) of the new solution?
2. If 17.0 mL of the stock solution are diluted to obtain a 0.179 M potassium carbonate 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 212 mL of a 0.228 M potassium carbonate solution.
4. Determine the volume of water (in mL) that was added to 29.7 mL of the stock solution to prepare a 0.839 M potassium carbonate solution.
Explanation / Answer
The following questions all refer to an aqueous 7.11 M potassium carbonate stock solution.
1. If 19.3 mL of the stock solution are diluted to 858 mL, what is the molarity (in mol/L) of the new solution?
M1 = 7.11 M , V1 = 19.3 , M2 = ? , V2 = 858 ml
M1 V1 = M2 V2
7.11 x 19.3 = M2 x 858
M2 = 0.160 M
new solution =0.160 M
2. If 17.0 mL of the stock solution are diluted to obtain a 0.179 M potassium carbonate solution, what is the volume (in mL) of the new solution?
7.11 x 17 = 0.179 x V2
V2 = 675 ml
new solution volume = 675 ml
3. Determine the volume (in mL) of the original stock solution that must be used to prepare 212 mL of a 0.228 M potassium carbonate solution.
7.11 x V1 = 212 x 0.228
V1 = 6.80 ml
4. Determine the volume of water (in mL) that was added to 29.7 mL of the stock solution to prepare a 0.839 M potassium carbonate solution.
29.7 x 7.11 = 0.839 x V2
V2 = 251.7 ml
volume of water added = V2 -V2 = 251.7 - 29.7 = 222 ml
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