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Chemistry Question... Due date is quickly approaching & I have been trying to figure this out with no such luck... PLEASE HELP?
How many total ions are present in 20.0 mL of a 1.50 M solution of Ca(NO3)2?
I remember that M= moles/Liters & 1 L= 1000 mL so 20.0 mL=0.02 L but I don't know where to go from there...
2 Answers
- KennyBLv 75 years ago
First, note that this molecule breaks into three ions - one Ca+2 and two NO3-2
Second, volume times molarity gives moles - 20.0 mL x 1.50 M is 30 millimoles.
3 x 10^-2 moles x 6.02 x 10^23 molecules/mole x 3 ions/molecule is about 5.4 x 10^22 ions.
Check my math (I just SWAGed it) and report the answer to three significant figures.
- 5 years ago
Solution: Find the number of moles and multiply by Avagadro's Number.
M= moles/Liters
moles= M*Liters
moles= (1.5M)(.02L)= .03 moles of Ca(NO3)2
Now convert from moles to molecules (ions in this case):
.03 x (6.02x10^23) = 1.806x10^22 molecules of Calcium Nitrate
Now when dissolved it will be Ca+2 and NO3- ***NOTE We have 2 nitrate ions for every 1 Calcium*
Multiply the number of molecules by the number of ions:
(1.806x10^22)x3 = 5.418 x 10^22 ions in the solution
Source(s): Chemistry Major