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1.575 g of oxalic acid (CO OH)(2).xH(2)O...

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.

Text Solution

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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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Knowledge Check

  • 1.575 of oxalic acid (COOH)_2.xH_2O are dissolved in water and the volume made upto 250 ml. On titration 16.68 ml of this solution requires 25 ml of N//15 NaOH solution for complete neutralization. Calculate x.

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    3
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  • 0.7 g "of" Na_(2)CO_(3).xH_(2)O were dissolved in water and the volume was made to 100 mL, 20 mL of this solution required 19.8 mL "of" N//10 HCl for complete neutralization. The value of x is:

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