Determine the molarity of solution of 25 gm oxalic acid (dissolved in one litre) solution?
class 11 th chemistry question.
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Normality : 0.04N
Steps:
1) Oxalic Acid: C2H2O4.(2H2O )
Molar Mass : 126g
Given Mass of Oxalic acid /solute = 0.63g
no. of moles of solute = 0.63/126 = 0.5 *10^(-2)
Volume of solution, V = 250cm^3 = 0.25L
=> Molarity, M
= no. of moles of solute / Volume of solution
Hence,Molarity = 0.02M
2) Since, Basicity / Valency Factor of acid is 2 .(Two ionisable H)
=> Normality =Molarity * Valency Factor
= 0.02M * 2 = 0.04N.
3) For Molality, Data is insufficient.
As We can't get Weight of solvent in solution from required data.
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