Solving for Volume G.5 a)

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Michael Du 1E
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Joined: Sun Sep 22, 2019 12:16 am

Solving for Volume G.5 a)

Postby Michael Du 1E » Tue Oct 01, 2019 10:37 pm

Can anyone please explain to me why, while solving for the volume, using the equation of mols solute/ molarity, the moles of Na2CO3 and moles of Na were taken into account and multiplied/ divided?

Vanessa Chuang 4F
Posts: 51
Joined: Sat Aug 24, 2019 12:18 am

Re: Solving for Volume G.5 a)

Postby Vanessa Chuang 4F » Tue Oct 01, 2019 11:04 pm

When you solve for volume, you are dividing moles of Na over molarity of Na2CO3 (mol Na/molarity Na2CO3). However the question is asking you to solve for the volume of Na. Molarity is expressed as mol/L so the Na2CO3 molarity can also be written as mol/L Na2CO3. You would then convert the moles of Na2CO3 into moles of Na (using ratio of Na per Na2CO3 -->which is 2 mols of Na). This would then give you moles of Na over liters of solution, which is what the answer is looking for. Hope this helps!

Juliet Stephenson 4E
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Joined: Wed Sep 18, 2019 12:21 am

Re: Solving for Volume G.5 a)

Postby Juliet Stephenson 4E » Tue Oct 01, 2019 11:15 pm

So this problem is asking us to determine the volume of the solution should be transferred into a flask in order to obtain 2.15 mmol of Na+. Importantly, this is not a dilution problem. Therefore, we won't be using the M1V1=M2V2 equation. The molarity of the solution won't be changing when we transfer it into a flask. By dividing the mass of Na2CO3 (2.111 g) by the molar mass (105.55 g.mol), we calculate that there are .0199 moles of Na2CO3 in the solution. Since we are attempting to find the millimoles of Na+, we'll multiply this quantity by 2 (from NA2). We now know that we have .0398 moles of Na+ in the solution. Therefore the molarity of this solution with regards to Na+ is (.0398 moles/.25L) or .1593 M. Since Molarity has units moles/liter, if we divide by the number of moles, we will be left with the liters of solution necessary. (.1592M/.00215M Na+)=.0135 L.

Hope this helps! Let me know if there is anything I can clarify!

Bita Ghanei 1F
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Joined: Thu Feb 28, 2019 12:15 am

Re: Solving for Volume G.5 a)

Postby Bita Ghanei 1F » Tue Oct 01, 2019 11:22 pm

Juliet Stephenson 4E wrote:So this problem is asking us to determine the volume of the solution should be transferred into a flask in order to obtain 2.15 mmol of Na+. Importantly, this is not a dilution problem. Therefore, we won't be using the M1V1=M2V2 equation. The molarity of the solution won't be changing when we transfer it into a flask. By dividing the mass of Na2CO3 (2.111 g) by the molar mass (105.55 g.mol), we calculate that there are .0199 moles of Na2CO3 in the solution. Since we are attempting to find the millimoles of Na+, we'll multiply this quantity by 2 (from NA2). We now know that we have .0398 moles of Na+ in the solution. Therefore the molarity of this solution with regards to Na+ is (.0398 moles/.25L) or .1593 M. Since Molarity has units moles/liter, if we divide by the number of moles, we will be left with the liters of solution necessary. (.1592M/.00215M Na+)=.0135 L.

Hope this helps! Let me know if there is anything I can clarify!


Hi there! I also needed help with this problem. Thank you for answering!
I was able to follow along until I got to the last line. Where did you get 0.00215M Na+ ?

Natalie Wang 1B
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Re: Solving for Volume G.5 a)

Postby Natalie Wang 1B » Tue Oct 01, 2019 11:30 pm

Sodium Carbonate has two sodium ions, so the ratio needs to be taken into account. When the problem asks you to find the volume for 2.15 mM of Na+, you need to convert 2.15 mM of Na+ into the amount of moles for sodium carbonate. To do that, you would need to show how 1 mol of sodium carbonate has 2 mols of Na+. I hope that explanation makes sense?

Juliet Stephenson 4E
Posts: 100
Joined: Wed Sep 18, 2019 12:21 am

Re: Solving for Volume G.5 a)

Postby Juliet Stephenson 4E » Tue Oct 01, 2019 11:32 pm

I had to google this before doing the problem, but mmol stands for millimoles, aka 1/1000 of a mole. So in order to keep all the units consistent, I converted the volume given in a) (2.15 mmol Na+) into moles by dividing by 1000 --> .00215 M Na+. Hope that makes sense!

William Francis 2E
Posts: 101
Joined: Wed Sep 18, 2019 12:20 am

Re: Solving for Volume G.5 a)

Postby William Francis 2E » Wed Oct 02, 2019 6:20 pm

Juliet Stephenson 4E wrote:So this problem is asking us to determine the volume of the solution should be transferred into a flask in order to obtain 2.15 mmol of Na+. Importantly, this is not a dilution problem. Therefore, we won't be using the M1V1=M2V2 equation. The molarity of the solution won't be changing when we transfer it into a flask. By dividing the mass of Na2CO3 (2.111 g) by the molar mass (105.55 g.mol), we calculate that there are .0199 moles of Na2CO3 in the solution. Since we are attempting to find the millimoles of Na+, we'll multiply this quantity by 2 (from NA2). We now know that we have .0398 moles of Na+ in the solution. Therefore the molarity of this solution with regards to Na+ is (.0398 moles/.25L) or .1593 M. Since Molarity has units moles/liter, if we divide by the number of moles, we will be left with the liters of solution necessary. (.1592M/.00215M Na+)=.0135 L.

Hope this helps! Let me know if there is anything I can clarify!

Juliet,
I believe you may have made a mistake in your explanation of the final step of this problem although I agree that the answer is 0.0135 L. 0.1592 divided by 0.00215 equals 74.05. I believe this last step to be an error because dividing a molarity (mol/L) by moles (mol) results in (L^-1) as the end unit rather than liters (L) as is desired. I look at the last step of the problem this way: (0.0398mol/.25L) is a ratio that can be set equivalent to the ratio of (0.00215mol/xL) where x is the desired number of liters of solution to be solved for. Hopefully that all makes sense! If not, let me know. I'm always ready to receive some new nubs of knowledge.
Will

Angela Prince 1J
Posts: 102
Joined: Sat Aug 24, 2019 12:17 am

Re: Solving for Volume G.5 a)

Postby Angela Prince 1J » Thu Oct 03, 2019 1:31 pm

William Francis 3C wrote:
Juliet Stephenson 4E wrote:So this problem is asking us to determine the volume of the solution should be transferred into a flask in order to obtain 2.15 mmol of Na+. Importantly, this is not a dilution problem. Therefore, we won't be using the M1V1=M2V2 equation. The molarity of the solution won't be changing when we transfer it into a flask. By dividing the mass of Na2CO3 (2.111 g) by the molar mass (105.55 g.mol), we calculate that there are .0199 moles of Na2CO3 in the solution. Since we are attempting to find the millimoles of Na+, we'll multiply this quantity by 2 (from NA2). We now know that we have .0398 moles of Na+ in the solution. Therefore the molarity of this solution with regards to Na+ is (.0398 moles/.25L) or .1593 M. Since Molarity has units moles/liter, if we divide by the number of moles, we will be left with the liters of solution necessary. (.1592M/.00215M Na+)=.0135 L.

Hope this helps! Let me know if there is anything I can clarify!

Juliet,
I believe you may have made a mistake in your explanation of the final step of this problem although I agree that the answer is 0.0135 L. 0.1592 divided by 0.00215 equals 74.05. I believe this last step to be an error because dividing a molarity (mol/L) by moles (mol) results in (L^-1) as the end unit rather than liters (L) as is desired. I look at the last step of the problem this way: (0.0398mol/.25L) is a ratio that can be set equivalent to the ratio of (0.00215mol/xL) where x is the desired number of liters of solution to be solved for. Hopefully that all makes sense! If not, let me know. I'm always ready to receive some new nubs of knowledge.
Will


I was confused by the 0.1592mol.L-1/.00215mol, so the way you expained it makes much more sense, Will. However, when I tried solving for x, I got 0.00135L instead of 0.0135L. Do you know what I could have done wrong?

William Francis 2E
Posts: 101
Joined: Wed Sep 18, 2019 12:20 am

Re: Solving for Volume G.5 a)

Postby William Francis 2E » Tue Oct 08, 2019 11:29 pm

To solve (0.0398mol/0.25L)=(0.00215mol/xL) for x, I would start by multiplying each side of the equation by x. The equation then becomes (0.0398x/0.25)=(0.00215). From here, I would multiply each side of the equation by 0.25, then divide each side by 0.0398 to get x=0.0135 as the final answer. Since you were off by a multiple of ten, I think you may have just input a number into your calculator with the decimal in the wrong spot or some small error like that at some point when solving for x. I hope that helps!


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