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

Answer

`6.02 xx 10^(21)` molecules.

Explanation

Gram equivalent of oxalic acid ` = (0.2 xx 100)/(1000) = 0.02`
Mass of oxalic acid `= 0.02 xx 63 = 1.26g`
126 g of oxalic acid contain ` = 6.02 xx 10^(23)` molecules .
1.26 g oxalic acid contain `= (6.02 xx 10^(23) xx 1.26)/(126)`
` = 6.02 xx 10^(21) ` molecules