# CHEMISTRY

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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 -

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 -

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 =(

• CHEMISTRY -

Well, you have to read the question carefully. It says aqueous solution of sodium ions, not sodium carbonate. So you have to divide 2.65g by 2. We use two in this question because there are 2 moles of Na in 1 mole of Na2CO3.

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