`0.63 g` of diabasic acid was dissolved in water. The volume of the solution was made `100 mL`. `20 mL` of this acid solution required `10 mL` of `N//5 NaOH` solution. The molecular mass of acid is:
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`underset(("Acid"))(N_(1)V_(1))-=underset(("NaOH))(N_(2)V_(2))` `N_(1)xx20=(1)/(5)xx10` `N_(1)=(1)/(5)xx(10)/(20)=(1)/(10)` Strength of the acid solution=Eq. mass of the acid `xx` Normality `Exx(1)/(10)=(E )/(10) g//L` Mass of acid in 100 mL of the solution `=(E )/(10)xx(100)/(1000)=(E )/(100)` Mass of acid in 100 mL of the solution =0.63 g (given) So, `(E )/(100)=0.63` or E=63 Mol. mass=Basicity `xx` Eq. mass `=2xx63=126`
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