Help on HW: Fundamentals G5

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Angela Prince 1J
Posts: 102
Joined: Sat Aug 24, 2019 12:17 am

Help on HW: Fundamentals G5

Postby Angela Prince 1J » Thu Oct 03, 2019 1:04 pm

Could someone please help me solve this problem? I got as far as converting 2.111g to moles (.020 mol) and finding the molarity (.8 mol.L-1). I don't know how to solve the rest. Thank you!
(p. F58)
G.5 A student prepared a solution of sodium carbonate by adding
2.111 g of the solid to a 250.0-mL volumetric flask and adding water
to the mark. Some of this solution was transferred to a buret. What
volume of solution should the student transfer into a flask to obtain
(a) 2.15 mmol Na+; (b) 4.98 mmol CO32-; (c) 50.0 mg Na2CO3?

Parker Smith
Posts: 102
Joined: Thu Jul 25, 2019 12:15 am

Re: Help on HW: Fundamentals G5

Postby Parker Smith » Thu Oct 03, 2019 1:12 pm

I'm going to use problem b for the example here. So how I began this problem: I figured out how many liters the .0796M sodium carbonate was in by using the moles of carbonate and the molarity of the solution. Use the equation "moles/volume=molarity." From there, I converted the liter measurement to milliliters for convenience. If this doesn't help, look at page 22-23 in the solutions book.

Chris Tai 1B
Posts: 102
Joined: Sat Aug 24, 2019 12:16 am

Re: Help on HW: Fundamentals G5

Postby Chris Tai 1B » Thu Oct 03, 2019 4:33 pm

I would start by using the equation for molarity and go from there:
Molarity = number of mols/volume

for part a, we can find the numbers of mols of Na2CO3 by taking 2.111 grams and dividing it by the molar mass of Na2CO3, which is about 106. Hence, we get 2.111/106 = 0.01992 mol Na2CO3. To find the molarity of the solution, we divide 0.01992mol/0.25L to get 0.07968 M. This is our concentration of Na2CO3 in the mixture.

I like to convert any mL or mmol to L or mol before continuing, so we can convert 2.15 mmol Na+ to 0.00215 mol Na+.

Next, we want to find the number of moles of Na+ that might be present given a certain amount of Na2CO3; since there are two Na+ ions for every Na2CO3 molecule, we only need 0.00215mol/2 = 0.001075 mol of Na2CO3 to find the exact volume it takes to get this amount of Na+ moles. Hence, we have our values:
Molarity = 0.07968 mol/L
mols Na2CO3 to get 0.00215 mol Na+ = 0.001075 mol

Plug in to solve: 0.07968 mol/L = 0.001075 mol/ V, V = 0.01349mL or 13.49 L.

Angela Prince 1J
Posts: 102
Joined: Sat Aug 24, 2019 12:17 am

Re: Help on HW: Fundamentals G5

Postby Angela Prince 1J » Thu Oct 03, 2019 9:12 pm

Chris Tai 1L wrote:I would start by using the equation for molarity and go from there:
Molarity = number of mols/volume

for part a, we can find the numbers of mols of Na2CO3 by taking 2.111 grams and dividing it by the molar mass of Na2CO3, which is about 106. Hence, we get 2.111/106 = 0.01992 mol Na2CO3. To find the molarity of the solution, we divide 0.01992mol/0.25L to get 0.07968 M. This is our concentration of Na2CO3 in the mixture.

I like to convert any mL or mmol to L or mol before continuing, so we can convert 2.15 mmol Na+ to 0.00215 mol Na+.

Next, we want to find the number of moles of Na+ that might be present given a certain amount of Na2CO3; since there are two Na+ ions for every Na2CO3 molecule, we only need 0.00215mol/2 = 0.001075 mol of Na2CO3 to find the exact volume it takes to get this amount of Na+ moles. Hence, we have our values:
Molarity = 0.07968 mol/L
mols Na2CO3 to get 0.00215 mol Na+ = 0.001075 mol

Plug in to solve: 0.07968 mol/L = 0.001075 mol/ V, V = 0.01349mL or 13.49 L.


Ok, thank you! That makes a lot more sense!
I was able to figure out the rest of the problem using your explanation :)


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