posted by STACY on .
how many grams of sodium carbonate are needed to prepare 0.250 L of an 0.100 M aqueous solution of sodium ions?
well normally you go moles=grams/molar mass than M= moles/L
but when looking for grams? what do you do........multiple .250 by .100 than divide by the molar mass of 105.988??
CHEMISTRY - DrBob222, Monday, December 13, 2010 at 8:51pm
You work the problem backwards.
moles = M x L.
Solve for moles. You got that far.
Then moles = grams/molar mass.
MULTIPLY (not divide) moles x molar mass.
CHEMISTRY - STACY, Tuesday, December 14, 2010 at 4:12pm
Moles= MxL (0.100M)(0.250L)=0.025
than multiply moles by molar mass
So my answer is 2.65g but the book says it's 1.32g? What am I overlooking.....man I hate chemistry =(
Chalk this one up to your book. I didn't read the question very well; in fact, I had to read it 3-4 times today before I caught why we were having trouble. The book answer is correct. 2.65 g Na2CO3 will make 250 mL of a 0.1M solution of Na2CO3. However, that isn't what the question asked. The question was to prepare 0.250L of a 0.1M aqueous solution of SODIUM IONS. So we have 2.65 grams Na2CO3 to prepare 250 mL of a 0.1 M Na2CO3 soln, and we want 0.1 M Na^+ (and there are two Na^+ in 1 molecule Na2CO3) so we need to divide our 2.65 g Na2CO3 by 2 to get where we want to go. Sorry about that. I don't always read these problems carefully enough. Thanks for bringing it back so we could get to the root of the problem.
0.025 mol Na^+
***for every one mole of Na2CO3, there are two moles of Na^+
so you'd have to divide that value by 2! and then multiply by the molar mass