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Question : 133 of 160
Marks:
+1,
-0
Solution:
Volume of solution
=250cm3=250‌mL =250×10−3L Mass of oxalic acid
=0.63g Molar mass of oxalic acid
=126 Moles
==5×10−3 moles
Molarity
=| Number‌of‌moles‌of‌solution |
| Volume‌of‌solution‌in‌litre |
M==0.02M Normality
= Basicity × Molarity
Basicity of oxalic acid
=2 N=2×0.02=0.04N
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