who likes doing other people's homework? :] (if you know chem?)

[len56]

I need the procedure for how to do this?

Our teacher will specify the volume and concentration (in mol/L) of sodium carbonate solution he wants us to make. We'll get sodium carbonate powder and distilled water to make our solution (and the appropriate tools/beakers like: electronic balance, powder funnel, etc.). What's the procedure to get the desired concentration (in mol/L) of the solution (with the specified volume)?

thanks if anyone is willing to help!

[ChineseGuy]

This sounds like highschool chemistry

[Grauss]

Molarity can only be determined within an aqueous solution, what you just stated is impossible. Sodium Carbonate in solid state has no molarity. However, if you simply put say 100.00g of Sodium Carbonate in 100mL of water then you would be able to find molarity quite quickly. The molecular mass of Sodium Carbonate is 105.98g which is equal to one mole of Sodium Carbonate, 100.00/105.98=0.94357 moles of Sodium Carbonate divided by 100mL (0.1L) of water (0.94357/0.1=9.4357) would give you a molarity of 9.4357. Now, all you'd have to do is plug it into Molarity 1 * Volume 1 = Molarity 2 * Volume 2. 9.4357 * 100 = (the molarity your teacher wants) * (the volume your teacher wants)

I hope that all makes sense, it's kinda difficult to explain over writing. And yes, that's definitely 1st year chem in high school lol

[len56]

Originally Posted by Grauss

Molarity can only be determined within an aqueous solution, what you just stated is impossible. Sodium Carbonate in solid state has no molarity. However, if you simply put say 100.00g of Sodium Carbonate in 100mL of water then you would be able to find molarity quite quickly. The molecular mass of Sodium Carbonate is 105.98g which is equal to one mole of Sodium Carbonate, 100.00/105.98=0.94357 moles of Sodium Carbonate divided by 100mL (0.1L) of water (0.94357/0.1=9.4357) would give you a molarity of 9.4357. Now, all you'd have to do is plug it into Molarity 1 * Volume 1 = Molarity 2 * Volume 2. 9.4357 * 100 = (the molarity your teacher wants) * (the volume your teacher wants)

I hope that all makes sense, it's kinda difficult to explain over writing. And yes, that's definitely 1st year chem in high school lol

hell no! it's grade 11 chemistry

yup, I'm sorry, I left that part out, we're supposed to make a 'Standard Sodium Carbonate Solution' with the given materials (water, sodium carbonate powder)

THANKS FOR THE HELP

[len56]

Originally Posted by Grauss

Molarity can only be determined within an aqueous solution, what you just stated is impossible. Sodium Carbonate in solid state has no molarity. However, if you simply put say 100.00g of Sodium Carbonate in 100mL of water then you would be able to find molarity quite quickly. The molecular mass of Sodium Carbonate is 105.98g which is equal to one mole of Sodium Carbonate, 100.00/105.98=0.94357 moles of Sodium Carbonate divided by 100mL (0.1L) of water (0.94357/0.1=9.4357) would give you a molarity of 9.4357. Now, all you'd have to do is plug it into Molarity 1 * Volume 1 = Molarity 2 * Volume 2. 9.4357 * 100 = (the molarity your teacher wants) * (the volume your teacher wants)

I hope that all makes sense, it's kinda difficult to explain over writing. And yes, that's definitely 1st year chem in high school lol

is that first half of the equation going to be the same no matter what?

[whyrms]

It's really easy since your sodium carbonate is in solid, we won't have to worry about it changing the volume. So all you'll have to do is weight out the specified amount of sodium carbonate then mix it with a specifide volume of water. The only trick to it is that the amount of solid is in mol, so you'll need to convert moles to grams, using the atomic weight which is written on the bottle (depending on what form you have 105.9884 g/mol (anhydrous) 124.00 g/mol (monohydrate) 286.14 g/mol (decahydrate)). The other problem you'll have to watch out for is how much of the solid you'll need. If he wants 0.5L of a 2 Moles/Litter solution, obviously, if you were making 1 Liter of this solution you'll need 2 Moles, but since you'll only make 0.5L of this solution, you'll only need to add 1 Mole.


chyea!!!

[DevJ]

I had a question somewhat similar to this on my AP chem test on the free response portion a few years ago..a little more detailed though lol

[DevJ]

BTW, is it supposed to be more detailed? Are you in AP Chem? Because if so, your teacher may want a detailed explanation (for example, fill with distilled water to etched mark on flask, etc)

[jibaholic101]

FUCK CHEM!

[Grauss]

whyrms has a point. I calculated for the anhydrous (and no, the first part would not be the same, it depends on the mass of sodium carbonate you use as well as the volume of water you dissolve it in) as opposed to a hydrated sample (which is still different from an aqueous solution lol) and yeah, if you're in AP Chem then you're going to want a more detailed answer than what I gave. Basically, we need more info =)

[eVitO]

Lol, I remember AP chem.

Whyrms and Grauss have you covered w/ this.

[synclastica_86]

AP = free credit.... so easy yet makes freshman year so much more fun!!!!!!!!! AP FTMFW!!!!!!!!!!