A sample of sodium carbonate contains sodium also. 1.5 g of the sample is dissolved in water and volume raised to 250 mL. 25 mL of this solution requires 20 mL of `(N)/(10)H_(2)SO_(4)` solution for neutralisation. Calculate the percentage of sodium carbonate in the sample.
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Only `Na_(2)CO_(3)` will react with `H_(2)SO_(4)`. Applying `underset((Na_(2)CO_(3)))(N_(1)V_(1)=underset((H_(2)SO_(4)))(N_(2)V_(2))` `N_(1)xx25=20xx(1)/(10)` `N_(1)=(20)/(25xx10)=0.08` Eq. mass of `Na_(2)CO_(3)=("Mol.mass")/(2)=(106)/(2)=53` Mass of `Na_(2)CO_(3)` present in 250 mL 0.08 N solution `(NxxExxV)/(1000)=(0.08xx53xx250)/(1000)=1.06g` Percentage of `Na_(2)CO_(3)` in the mixture `=(1.06)/(1.50)xx100=70.67`
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