1.575 g of oxalic acid `(CO OH)_(2).xH_(2)O` are dissolved in water and the volume made up to 250 mL. On titration 16.68 mL of this soltuion requires 25 mL of `(N)/(15)NaOH` solution for complete neutralisation. Calculate x.
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25 mL of `(N)/(25)NaOH` solution `-=25 mL " of" (N)/(15)` oxalic acid solution Mass of oxalic acid present in 25 mL of `(N)/(15)` oxalic acid solution `=(NxxExxV)/(1000)=(1xx(90+18x)xx25)/(15xx2xx1000)` `=((90+18x))/(1200)g` Actually `((90+18x))/(1200)` g oxalic acid is present in 16.68 mL solution. 250 mL of the solution contains oxalic acid `=((90+18x)xx250)/(1200xx16.68)=1.575`(given) or `90+18x=(1.575xx1200xx16.68)/(250)=126` or 18=126-90=36 x=2
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