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Sodium carbonate was dissolved in 100 millilitre of water then reacted with excess calcium acetate 4.0 gram of precipitate was formed the concentration in mol/litre of the sodium carbonate is

Accepted Answer
1. Convert the mass of calcium acetate to moles

4.0 g Ca(CH3COO)2 * (1 mol Ca(CH3COO)2 / 158.17 g Ca(CH3COO)2) = 0.0253 mol Ca(CH3COO)2

2. Calculate the moles of sodium carbonate that reacted with calcium acetate

The reaction between sodium carbonate and calcium acetate is a 1:1 molar ratio.

So, 0.0253 mol Ca(CH3COO)2 * (1 mol Na2CO3 / 1 mol Ca(CH3COO)2) = 0.0253 mol Na2CO3

3. Calculate the concentration of sodium carbonate in mol/L

Concentration = moles of solute / volume of solution in liters

Concentration = 0.0253 mol / 0.1 L = 0.253 M

Therefore, the concentration of sodium carbonate is 0.253 mol/L.

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