Solving 0.005M Na2CO3 Chemistry Problem

[higherme]

[SOLVED] Chemistry Problem

The question is
Calculate the mass of pure Na2CO3 needed for the preparation of 250.00mL of a 0.005M standard carbonate solution.

when it says PURE Na2CO3, what molarity does it mean? is it 1M??

my guess is to use M1V1 = M2V2 to solve
my final volume is 250mL
final concentration is 0.005 M
and if the pure Na2CO3 means 1M then that would be what i would be using as my stock right? then i would solve for the volume of that stock i need. After that I would use the density of Na2CO3 to find the mass

is that right?

 

[chemisttree]

Find out how many moles of sodium carbonate you will need to make up this solution.
Hint: #moles = molarity(moles/L)/volume(L)

 

[higherme]

so I figured that I will need 0.005M/0.25L = 0.02 mol of carbonate

now I use the molar mass of sodium carbonate to find the mass ??

 

[eli64]

yes, correct.

 

[higherme]

Thanks!

 

[higherme]

wait.. i thought molarity is found by (mol/L) x L?
and not (mol/L) / L


?

 

[higherme]

i mean moles is found by (mol/L) x L

 

[chemisttree]

What is your question?

 

[higherme]

before, you said that #moles = molarity(moles/L)/volume(L)

i thought #moles = molarity(moles/L) x volume(L)??

 

[eli64]

higherme you are correct, Molarity = mol/L;
so mol = M x L

probably a typo in the reply before. then multiply by Molar mass to get grams

 

[chemisttree]

Yes higherme, you are correct. Definitely a 'typo' (screwup) on my part...