"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 the solution was transferred to a buret. What volume should the student transfer into a flask to obtain 2.15 mmol Na+?"
I understand that you have to convert the grams of Na2Co3 to moles and then divide by the given volume to obtain the molarity, and I got the right answer. But since you have to multiply by the molar ratio to get from moles of Na2Co3 to moles of Na+ how come you can only multiply in the last step when you're already finding the volume as opposed to when you're first converting the grams of Na2Co3? Please let me know if that needed clarification because it sounds kind of wordy.
G5
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Re: G5
-1st step involves calculating the moles of Na2CO3
2.111gNa2CO3
_____________________ = .01992 mol NaCo3
105.991 g/mol Na2CO3
-2nd step
.01992 mol/.250L=.0769 mol/L or about .080mol/L
(a) Convert mmol to mol .002015 mol Na/0.16 mol/L = 0.013 L Na x 1000ml = 13 mL Na is needed to transfer
2.111gNa2CO3
_____________________ = .01992 mol NaCo3
105.991 g/mol Na2CO3
-2nd step
.01992 mol/.250L=.0769 mol/L or about .080mol/L
(a) Convert mmol to mol .002015 mol Na/0.16 mol/L = 0.013 L Na x 1000ml = 13 mL Na is needed to transfer
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Re: G5
I also don't understand this. Intuitively, for every one mole of Na2CO3, shouldn't there be 2 moles of Na? And why is the multiplication done in the last step?
2.11g Na2Co3= 0.0199 moles Na2Co3
(1 mole Na2Co3)*(2 mol Na/1 mole Na2Co3)= 0.0398M Na Solution
(0.0398)(x)=(2.15*10^3)(0.25)
x= 3.4 mL
Can someone explain please?
2.11g Na2Co3= 0.0199 moles Na2Co3
(1 mole Na2Co3)*(2 mol Na/1 mole Na2Co3)= 0.0398M Na Solution
(0.0398)(x)=(2.15*10^3)(0.25)
x= 3.4 mL
Can someone explain please?
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- Joined: Fri Aug 30, 2019 12:17 am
Re: G5
Hi Hussain!
I was initially confused at first as well; it does make sense to have 2 moles of Na for every mole of Na2CO3.
I think when you multiplied this, it should be 0.0398 moles of Na rather than Molarity of the solution:
(1 mole Na2Co3)*(2 mol Na/1 mole Na2Co3) = 0.0398 moles of Na
Then, you would divide that by 0.250L to get a 0.1592M Na Solution
Instead of multiplying, I used the equation: Volume = Moles/Molarity:
Volume = (2.15*10^-3 moles) / 0.1592M = 13.5 mL
The Equation M1V1 = M2V2 is used for dilutions, so I don't think it would be the most convenient to use in this problem.
Hope this helps you!!
I was initially confused at first as well; it does make sense to have 2 moles of Na for every mole of Na2CO3.
I think when you multiplied this, it should be 0.0398 moles of Na rather than Molarity of the solution:
(1 mole Na2Co3)*(2 mol Na/1 mole Na2Co3) = 0.0398 moles of Na
Then, you would divide that by 0.250L to get a 0.1592M Na Solution
Instead of multiplying, I used the equation: Volume = Moles/Molarity:
Volume = (2.15*10^-3 moles) / 0.1592M = 13.5 mL
The Equation M1V1 = M2V2 is used for dilutions, so I don't think it would be the most convenient to use in this problem.
Hope this helps you!!
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- Joined: Sat Aug 17, 2019 12:15 am
Re: G5
To solve this, because the solutions are equimolar, I set up the proportion:

If you cross multiply to solve for volume, you'll get V=13.5ml
If you cross multiply to solve for volume, you'll get V=13.5ml
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- Joined: Sat Sep 07, 2019 12:16 am
Re: G5
Naomi 3G wrote:Hi,
I am also confused why in the solutions book for a they multiply .07977 by 2 in the denominator of the equation
They multiplied 0.07977 by 2 in the denominator because the 2 is from the 2 moles of Na+ that make up the 1 mole of Na2CO3. Hope this clarifies things!
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