## Test 1, Q1 [ENDORSED]

Brynne Keyser 1B
Posts: 31
Joined: Fri Apr 06, 2018 11:03 am

### Test 1, Q1

If someone could walk me through this problem, it would be much appreciated!

A) Lithium carbonate, Li2CO3, is a salt that is used for the treatment of bipolar disorder. A solution of lithium carbonate is prepared by dissolving 2.0 g of salt to 1.00 L water. What is the concentration of lithium ions for the above solution?

B) The above solution is diluted with 1.00 L of 0.100 M of lithium chloride, LiCl, solution. What is the concentration of lithium ions after the dilution?

Alejandro Salazar 1D
Posts: 31
Joined: Fri Sep 29, 2017 7:05 am

### Re: Test 1, Q1

For part A, did you get M= 73.9. I'm not sure if I did correctly either. I simply used M= n(number of moles)/V(Liters)

Emely Reyna 1F
Posts: 30
Joined: Fri Apr 06, 2018 11:04 am

### Re: Test 1, Q1  [ENDORSED]

For part B to find the concentration of the lithium ions you find the moles from the previous question and add the 0.100 mols from the new solution. You then divide by the total volume, which in this problem would be 2.00 L.

SammiOrsini_1B
Posts: 45
Joined: Fri Apr 06, 2018 11:04 am

### Re: Test 1, Q1

For part A of the question I got .054 M. This number is extremely different from the one above. If anyone knows how to do this problem could they please walk me through it. I found the molar mass of the salt and got 73.892 g/mol and I found that there are .054 mol of Li in the solution. I then divided the .o54 by 1 L and got .054 M.

Chem_Mod
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### Re: Test 1, Q1

Make sure to find the number of moles of Li+ for part A, which is twice the number of moles of Li2CO3