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 is

Options

A. 12.5 g
B. 1.25 g
C. 0.125 g
D. 125 g

Answer

B

Explanation

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`