G.5

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Jeril Joseph 1B
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G.5

Postby Jeril Joseph 1B » Tue Jun 25, 2019 4:13 pm

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 mol Na+ ...?

Can someone walk me through the process of solving this problem because I don't understand it. Thank you.

David Zhang 1B
Posts: 48
Joined: Mon Jun 17, 2019 7:23 am

Re: G.5

Postby David Zhang 1B » Tue Jun 25, 2019 8:22 pm

You first have to convert the 2.111g of the solid sodium carbonate into mmol of Na+. I used dimensional analysis to do the conversion. Then you can take that answer to set up a proportion since you know that n/v = n/v. You know the values of mmol and volume for the initial solution and the mmol for the final solution and can then solve for the volume.

Daniel Kim 1D
Posts: 38
Joined: Wed Oct 03, 2018 12:17 am

Re: G.5

Postby Daniel Kim 1D » Wed Jun 26, 2019 7:33 pm

When I saw this problem for the first time, I thought i had to use M1V1 = M2V2, however looking closer at the problem you don’t need it. First you want to find the molarity of Na2CO3 which is .08 M Na2CO3. Then you change the moms of Na to mol. You want to find the volume (in liters) so using dimensional analysis, you divide the mol of Na by the molarity so that you can cancel out the moles and have only liters in your answer at the end. Same for the other 2, you would want to convert everything into moles so you can compare them.

hannabarlow1A
Posts: 37
Joined: Mon Jun 17, 2019 7:23 am

Re: G.5

Postby hannabarlow1A » Wed Jun 26, 2019 8:20 pm

For part A:

First, I calculated the molar mass of the compound, which is 105.99 g/mol. Next, to convert this value to moles, I calculated 2.111 g divided by the molar mass (g/mol) so that grams cancel out and we are just left with moles. Now that we have the number of moles (.0199 moles), we can divide this value by volume (.250 liters) to calculate molarity. 0.0199 moles divided by 0.25 L is 0.0797 moles per Liter. Since the problem is asking about sodium ions, we multiply this molarity by 2, because there are 2 sodium ions in the molecule. Now, we can finally use the formula v=n/M to solve for v. V= (2.15 x 10^-3) divided by 0.159. Solving this, we find that the volume we want is 0.0135 L.


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