Anyone do G15 A?

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janeane Kim4G
Posts: 31
Joined: Fri Sep 28, 2018 12:28 am

Anyone do G15 A?

Postby janeane Kim4G » Tue Oct 02, 2018 2:48 pm

Heya back with a molarity q, what volume of .778 M Na2CO3(aq) shouldbe diluted to 150.0 mL of water to make the concentration 0.0234 M NaCO3(aq)? How do you start/ approach problems like this?

Kenan Kherallah 2C
Posts: 78
Joined: Fri Sep 28, 2018 12:17 am

Re: Anyone do G15 A?

Postby Kenan Kherallah 2C » Tue Oct 02, 2018 3:00 pm


So I would start by getting the moles of Na2CO3 needed. You would do this by 0.15 * 0.0234 M to get 0.00351.
By doing this you then have to solve for the volume of the concentration 0.778M. 0.00351/0.778 = 0.00451 L

Jessica Chen 1F
Posts: 61
Joined: Fri Sep 28, 2018 12:17 am

Re: Anyone do G15 A?

Postby Jessica Chen 1F » Tue Oct 02, 2018 3:01 pm


So you can first list out the basic formulas needed here:
M(molarity) = mol/volume(L)
MiVi=MfVf (because the # of moles are constant; i means initial, f means final)

You can then list out the given information:
Mi = 0.778 M
Vi = ?
Mf = 0.0234 M
Vf = 150.0 mL

You just have to plug all the given values into the formula(don't forget to convert Vf into liters!) and solve for Vi.
It's going to look like this:
Vi = MfVf/Mi
Vi = (0.0234M)(0.1500L)/(0.778M)
Vi = 0.00451 L

Converting this answer back to mL, we get 4.51 mL.

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