Calculate the molality oF a solution containing 5.3gm of anhydrous Na2CO3 in 400g of water

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Solution

The molecular mass of Na₂CO₃= 106g
Mass of Na₂CO₃ = 5.3g
Mass of water = 400g = 0.4kg
No of moles of Na₂CO₃ ,(n)= Mass of Na₂CO₃/Molecular mass of Na₂CO₃ = 5.3/106 = 0.05
Molality = No of moles/ mass of solvent in kg = 0.05/.4 =0.125

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