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
Soooo
Moles= MxL (0.100M)(0.250L)=0.025
than multiply moles by molar mass
(0.025)(105.988)= 2.65g
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
To calculate the number of grams of sodium carbonate needed to prepare the solution, you can follow these steps:
1. Start with the given volume of the solution, which is 0.250 L.
2. Next, use the molarity of the solution, which is 0.100 M, to determine the number of moles of sodium ions in the solution. The formula for moles is Moles = M x L, where M is the molarity and L is the volume in liters.
So, Moles = 0.100 M x 0.250 L = 0.025 moles.
3. Now, use the molar mass of sodium carbonate to convert the moles of sodium ions to grams of sodium carbonate. The molar mass of sodium carbonate (Na2CO3) is 105.988 grams/mol.
Grams = Moles x Molar mass = 0.025 moles x 105.988 grams/mol = 2.65 grams.
According to these calculations, the answer is indeed 2.65 grams. However, you mentioned that the book states the answer is 1.32 grams. This suggests there may be an error in the given molarity or a misprint in the book.