How to find volume of solution w specific amount of Na2CO3?

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

How to find volume of solution w specific amount of Na2CO3?

Postby janeane Kim4G » Tue Oct 02, 2018 12:17 am

It’s G5 part c on the 7th edition for reference. If a student has a 250mL solution of 2.111g of sodium carbonate, what volume of the solution should be transferred into a separate flask to obtain 50.0 mg of Na2CO3?

ChathuriGunasekera1D
Posts: 78
Joined: Fri Sep 28, 2018 12:17 am

Re: How to find volume of solution w specific amount of Na2CO3?

Postby ChathuriGunasekera1D » Tue Oct 02, 2018 8:03 am

From part (a) and (b) we know that the original concentration of the sodium carbonate (2.111 g) is .07966 M. This comes from converting grams of sodium carbonate to moles, and then dividing that by the total volume in liters

(2.111 g) / (106 g) = 0.0199 mol Na2CO3 / 0.250 Liters = 0.07966 M Na2CO3

Then we use the amount they gave us and work backwards. I converted 50 mg of Na2CO3 to grams, then to moles using molar mass (106 g/mol), and then to liters using concentration.

(50 mg) / (1000 mg/g) = 0.05 g Na2CO 3
(0.05 g) / (106 g/mol) = 0.0004717 mol Na2CO3

We know that if the concentration of the original solution is 0.07966, then there are 0.07966 moles per every liter.

(0.0004717 mol) / (0.07966 mol/L) = 5.92 x 10^(-3) L

Matthew Tran 1H
Posts: 165
Joined: Fri Sep 28, 2018 12:16 am

Re: How to find volume of solution w specific amount of Na2CO3?

Postby Matthew Tran 1H » Tue Oct 02, 2018 10:42 pm

An easier way to do this is to set a proportion. The molarity doesn't change since you are not adding solute nor diluting the solution; thus you can set up the proportion M1 = M2. Using the definition of molarity, the proportion is equivalent to m1/V1 = m2/V2 (m = mass, V = volume; you can use mass instead of moles since the ratio of masses is the same as moles). In the problem you are given m1, V1, and m2, leaving you with one unknown, V2, which can easily be solved for. This saves you time because you don't have to even calculate the molarity or the molar mass.


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