A solution of oxalic acid (COOH)2.2H2O is prepared by dissloving 0.63g of acid in 250 cm3 of the solution . calculate the molarity of the solution.
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Moles of solute = Weight / Gram Molecular weight
= 0.63 g / 126 g/mol
= 0.005 mol
Volume of solution = 250 cm³ = 0.25 Litres
Molarity = Moles of solute / Volume of solution in litres
= 0.005 / 0.25
= 0.02
Molarity of solution is 0.02 M
= 0.63 g / 126 g/mol
= 0.005 mol
Volume of solution = 250 cm³ = 0.25 Litres
Molarity = Moles of solute / Volume of solution in litres
= 0.005 / 0.25
= 0.02
Molarity of solution is 0.02 M
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