chemistry solution prepare

Question

HOW TO PREPARE:

200 ml 0.5 M CH3COONa (you have solid CH3COONa.3H20) (M CH3COONa: 82.03 g/mol)

Answer 1

0.5 M CH3COONa means you want 0.5 moles of CH3COONa in 1 liter of solution.

Since we wish to prepare only 0.200 L (200 ml) of solution, we calculate moles needed as follows:

0.5 mols / L x 0.2 L = 0.1 moles of CH3COONa needed

Since we have only the trihydrate (CH3COONa-3H2O) available, we use the molar mass of this to calculate grams needed in order to provide the desired 0.1 moles of CH3COONa

molar mass CH3COONa-3H2O = 136 g / mol

0.1 mol CH3NOONa-3H2O provides 0.1 mol CH3COONa

0.1 mol x 136 g / mole = 13.6 g

So, we need 13.6 g but we round to 14 g since 0.5 M has only 1 significant figure.

Procedure:

Weigh out 14 g of CH3COONa-3H2O and dissolve in sufficient solvent (H2O) to make a final volume of 200 mls.