Help with G5

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Helen Mejia 1I
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Joined: Fri Sep 28, 2018 12:24 am

Help with G5

Postby Helen Mejia 1I » Sat Oct 06, 2018 3:57 pm

Hi can someone explain how to do G5 (edition 7)

It says " A student prepared a solution of sodium carbonate by adding 2.111g 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) 2.15mmol Na+ (b) 4.98 mmol C0^-23 (the three is in the bottom of the -2) (c) 50.0mg Na2CO3?


Please and thank you.

JulieAljamal1E
Posts: 71
Joined: Fri Sep 28, 2018 12:24 am

Re: Help with G5

Postby JulieAljamal1E » Sat Oct 06, 2018 4:48 pm

You start by solving for the molarity of Sodium Carbonate by first dividing the given mass of Na2CO3 (2.111g) by its molar mass which is 105.99 g/mol to get grams into moles. Then you take the number of moles you just calculated (.01992 mol) and divide that by the given volume of 0.25 L (I converted the 250.0 mL to L). The quotient of # mols/volume will give you the molarity 0.07967 mol/L. Knowing this piece of information you can continue on to part A. Another critical piece of info. to know is that Sodium Carbonate plus water dissociates into two Na cations and one CO3 anion. You can discover this by balancing its chemical equation. Now knowing this you want to set up a ratio that will allow units to cancel and leave you with Liters because we’re looking for volume. So you can put (0.00215 mol Na+ (the number of moles the problem wants in the solution)/ 2 mole Na+) (which we know we have from the balanced equation) and multiply that by (1 mole Na2CO3/ 0.07967 mol/L). All the units cancel except L and you’re left with 1.35x10^-2 L for part A. If it helps, you should look at the units you’re trying to end with and the units of the given information or information you can solve for, and try to see how you can cancel out units to get the desired one (liters in this case). Part B is the same principal but with CO3, so you make a ratio of 0.00498 mol over the number of moles of CO3 which is 1 mole.

Lina Petrossian 1D
Posts: 77
Joined: Fri Oct 05, 2018 12:16 am

Re: Help with G5

Postby Lina Petrossian 1D » Sun Oct 07, 2018 1:06 pm

For problem G5, I got the following answers: a. 0.013 L, b. 0.06225 L, and c. 0.0059 L. Did anyone else get these answers?

daniella_knight1I
Posts: 57
Joined: Fri Sep 28, 2018 12:18 am

Re: Help with G5

Postby daniella_knight1I » Sun Oct 07, 2018 5:36 pm

Can you explain the steps you took to get those answers? I can point out and help with the mistakes you might've made.

Reese - Dis 1G
Posts: 31
Joined: Fri Sep 28, 2018 12:15 am

Re: Help with G5

Postby Reese - Dis 1G » Mon Oct 08, 2018 7:27 pm

Is there a specific formula we're supposed to use to solve this problem?

Albert Duong 4C
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Joined: Fri Sep 28, 2018 12:17 am
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Re: Help with G5

Postby Albert Duong 4C » Mon Oct 08, 2018 11:29 pm

Not really, it's stoichiometry, so it's cancelling out units using division and multiplication until we get the units we want. For example:
[0.00251 mol Na ] [1 mol Na2CO3] [ 1 L ]
-------------------------------------------------------------------------- = 1.35x10^-2 L
[2 mol Na ] [0.07967 mol Na2CO3]

Notice how we can divide or multiply away the units (I know, the "boxes" are garbage, but I find them helpful for organizing the data for stoichiometry), such as how we used the reciprocal of 0.07967 mol/L of Na2CO3 to divide away the moles of Na2CO3 so we're just left with liters. Hopefully this visual helps!


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