G.5  [ENDORSED]

Moderators: Chem_Mod, Chem_Admin

Future-Hope Vang 3L
Posts: 2
Joined: Fri Sep 29, 2017 7:06 am

G.5

Postby Future-Hope Vang 3L » Fri Oct 06, 2017 10:52 am

I got the molarity which is about 0.0796 M. What do I do after this?

Kevin Liu 3G
Posts: 28
Joined: Fri Sep 29, 2017 7:07 am

Re: G.5  [ENDORSED]

Postby Kevin Liu 3G » Fri Oct 06, 2017 11:52 am

After you get the M of NA2CO3, you have to figure out how many moles of NA+ there is. Since it says mmol it would be 2.15 x 10^-3 moles of Na2 but you want NA+ so you would divide (2.15 x 10^-3)/2 getting 1.075 x 10^3. Then you use the formula that V = n/M and you plug in the numbers and get 1.35 x 10^-3 L.

For part b, you would do the same thing. 4.98 mmol --> becomes 4.98 x 10^-3 mols
then doing the same thing you should get 6.25 x 10^-3 L

Future-Hope Vang 3L
Posts: 2
Joined: Fri Sep 29, 2017 7:06 am

Re: G.5

Postby Future-Hope Vang 3L » Sun Oct 08, 2017 9:13 pm

Thanks! I didn’t get the mmol and “x 10^-3” part so this really helped.


Return to “Molarity, Solutions, Dilutions”

Who is online

Users browsing this forum: No registered users and 2 guests