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??

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.

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 =(

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.

To calculate the grams of sodium carbonate needed, you can follow these steps:

1. Start by determining the number of moles of sodium ions required in the solution. This can be done by multiplying the desired molar concentration (0.100 M) by the volume of the solution (0.250 L).

Number of moles = Molar concentration × Volume
Number of moles = 0.100 M × 0.250 L

2. Once you have determined the number of moles, you can use the molar mass of sodium carbonate to convert moles to grams. The molar mass of sodium carbonate (Na2CO3) is 105.988 g/mol.

Grams = Number of moles × Molar mass
Grams = (0.100 M × 0.250 L) × 105.988 g/mol

3. Calculate the final result.

Grams = 0.025 g × 105.988 g/mol
Grams ≈ 2.65 g

Therefore, approximately 2.65 grams of sodium carbonate are required to prepare 0.250 L of a 0.100 M aqueous solution of sodium ions.