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March 31, 2015

March 31, 2015

Posted by **John** on Wednesday, December 15, 2010 at 8:18pm.

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.

- Chemistry (URGENT) -
**DrBob222**, Wednesday, December 15, 2010 at 8:45pmThe 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.

- Chemistry (URGENT) -
**John**, Wednesday, December 15, 2010 at 8:55pmOkay, but what does it mean "Calculate the molarity of the Co^2+ ion in each of the solutions 1-5"

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