G5

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Eden Breslauer-Friedman 2A
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G5

Postby Eden Breslauer-Friedman 2A » Thu Oct 08, 2020 9:59 am

How would you go about starting this problem?

G5. 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 mmol Na+; (b) 4.98 mmol CO32−;(c) 50.0mg Na2CO3?

Thank you!

Chem_Mod
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Re: G5

Postby Chem_Mod » Thu Oct 08, 2020 10:36 am

Begin this problem by converting the 2.111g to moles by dividing by the compound's molar mass. Next, divide by 0.2500 L (converting mL to L) in order to find the molarity (moles/L).

805377003
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Re: G5

Postby 805377003 » Thu Oct 08, 2020 11:03 am

I am also confused. After you find the molarity, what is the next step?

Marley Magee 3A
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Re: G5

Postby Marley Magee 3A » Thu Oct 08, 2020 11:27 am

This problem uses the equation c=n/v, which we can rearrange to v=n/c. "v" being the desired volume, "n" the number of moles, and "c" the concentration. So once you have the concentration of of Na2CO3 you convert it to the concentration of just Na in the 250 mL for part A. Once you have the concentration of Na you can plug it in as "c" with 0.00215 mol as "n" (which they gave you as 2.15 mmol in the problem). Then solve for the volume.
In the end, should look something like this:
0.00215 mol Na / 0.1594 mol Na*L^-1 = 1.349*10^-2 L

Tatyana Bonnet 2H
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Re: G5

Postby Tatyana Bonnet 2H » Thu Oct 08, 2020 11:30 am

After finding the molarity, you use it to find the molarity of the compound or element you are trying to find the volume for. For A )that would be Na+. Looking at the compound in the solution which is sodium carbonate Na2(CO3) you see that Na has a subscript of 2 so the amount of mols of the compound would be multiplied by two since one mol of Na2 contains 2 Na atoms in the molecule. This gives you the molarity of the Na.

(.0797 mol/L)x 2 = .159mol/L for Na

You would then convert the mmol to mol getting 0.00215 mol of Na

Using M=n/V you would rearrange it to get V=n/M and then plug in the variables accordingly

V=.00215mol/.159mol/L --> 0.0135L which is equal to 13.5 mL

Ellison Gonzales 1H
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Re: G5

Postby Ellison Gonzales 1H » Fri Oct 09, 2020 12:46 am

Regarding finding the volume of solution for b.) Co32-, does Co32- have only one mole? I don’t exactly remember how the 2- here effects the moles. Since we multiplied the molarity by 2 for Na2 for part a, I assumed we would multiply the molarity by 3 for Co32-, but the solution says otherwise.

arisawaters2D
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Re: G5

Postby arisawaters2D » Fri Oct 09, 2020 10:26 am

I'm still a little confused with this question. I understand how to get the molarity of Na2(CO3), but I'm struggling with where to go after that. For part a, how do you convert mmol to mol? Could someone explain how to do parts a,b, or c in depth? Thank you!

Marley Magee 3A
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Re: G5

Postby Marley Magee 3A » Fri Oct 09, 2020 11:00 am

a) The concentration of Na2CO3 should turn out to be around 0.07968 mol/L. Now we need to find the concentration of just Na from the concentration of Na2CO3. To do this, we know that there are 2 moles of Na per one mole of Na2CO3 (because the subscript of Na is 2). Therefore, we take the concentration of Na2CO3 and multiply it by 2, to get 0.1594 mol/L for Na.
(0.078968 mol Na2CO3 / 1L) * (2 mol Na / 1 mol Na2CO3) = 0.1594 mol Na/L

Then, we can plug it into the equation v = n/c, with "n" being the desired moles of Na that they give you in the problem, 2.15 mmol. The "m" stands for milli, which means it's 1/1000th of a liter. Therefore, we are looking for 0.00215 mol Na. The "c" is the concentration of Na we just found, 0.1594 mol/L. Plugging in we get:
v = 0.00215 mol Na / 0.1594 mol Na/L = 1.349*10^-2 L.
1.349*10^-2 L is the final answer.

Parts b and c go through the same procedure, just with different numbers. All parts of this problem use v=n/c.

arisawaters2D
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Re: G5

Postby arisawaters2D » Fri Oct 09, 2020 12:07 pm

Thank you so much Marley!!


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