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In a titration `20 cm^(3)` of 0.1 N oxalic acid solution requires `20 cm^(3)` of sodium hydroxide for complete neutralization. The mass of sodium hydroxide in `250 cm^(3)` solution isA. 12.5 gB. 1.25 gC. 0.125 gD. 125 g |
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Answer» Correct Answer - B Now `underset("Acid")(N_(1)V_(1))=underset("Base")(N_(2)V_(2))` `N xx 25=.1 xx 25` or `N_(1)=(0.1 xx 25)/(25)=0.1 N` Now `N=("Mass of solute")/("Eq mass")xx(1000)/("Volume in mL")` `0.1 =("Mass of solute")/(63)xx(1000)/(200)` `:. ` Mass of solute`=(63 xx0.1)/(5)=1.25g` |
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