Calculate the number of molecules of exalic acid `(H_(2)C_(2)O_(4)2H_(2)O)` in 100 mL of 0.2 N oxalic acid solution.

Explanation

Calculation of mass of oxlic acid
`" Normality of solution"=("Mass of oxalic acid"//"Equivalent mass")/("Volume of solution in litres")`
`"Normality mass of acid"= ("Molar mass")/("Basicity")=(2xx1+2xx12+4xx16+2xx18)/2=126/2`
`63 g equiv^(-1)`
Volume of solution =100mL=100/1000=0.1L
`(0.2"equiv"L^(-1))=("Mass of oxalic acid")/(63"gequiv"^(-1)xx(0.2L))`
`"Mass of oxalic acid"=(0.2equiv.L^(-1))xx(63 g equiv.^(-1))xx(0.1L)=1.26 g`
Calculation of no. of molecules of oxalic acid.
Molar mass of xalic acid =`126 g mol^(-1)`
Now, 126 gof exalic acid has molecules=`(N_(0)=6.022xx10^(23))`
1.26 g of oxalic acid has molecules=`(6.022xx10^(23))/((126g))xx(1.26g)=6.022xx10^(21)`