Asked by John
Calculate the molarity of the Co^2+ ion in each of the solutions 1-5. The new molarity is given by the molarity of the stock solution multiplied by the dilution factor.
Solution: Co(NO3)2
Original molarity: 0.160
Dilution #1: none
D#2: 12mL to 16mL
D#3: 8mL to 16mL
D#4: 4mL to 16mL
D#5: 2mL to 16mL
Please help I do not understand what I'm supposed to do. Thank you.
Solution: Co(NO3)2
Original molarity: 0.160
Dilution #1: none
D#2: 12mL to 16mL
D#3: 8mL to 16mL
D#4: 4mL to 16mL
D#5: 2mL to 16mL
Please help I do not understand what I'm supposed to do. Thank you.
Answers
Answered by
DrBob222
The problem is giving you 0.150M as the molarity and it tells you the new molarity will be the old one (0.160) x the dilution factor). After diluting, the first one will be 0.160 x (12mL/16mL)
The next one is 0.160 x (8 mL/16 mL).
However, from the way the problem is states I think it wants you to write the new molarity as the old molarity x the FACTOR. The dilution factor for the first one is (12/16) = 3/4; therefore, the first one will be
0.160 x (3/4), which of course is exactly the same answer as 0.160 x (12/16) but 3/4 is the dilution factor.
The next one is 0.160 x (8 mL/16 mL).
However, from the way the problem is states I think it wants you to write the new molarity as the old molarity x the FACTOR. The dilution factor for the first one is (12/16) = 3/4; therefore, the first one will be
0.160 x (3/4), which of course is exactly the same answer as 0.160 x (12/16) but 3/4 is the dilution factor.
Answered by
John
Okay, but what does it mean "Calculate the molarity of the Co^2+ ion in each of the solutions 1-5"
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