A aqueous solution of `NaCl` on electrolysis gives `H_(2)(g), Cl_(2)(g)`, and `NaOH` according to the reaction :
`2Cl^(-)(aq)+2H_(2)Orarr 2OH^(-)(aq)+H_(2)(g)+Cl_(2)(g)`
A direct current of 25A with `62%` efficiency is passed through 20 L NaCl solution (`20%` by weight). (3) What will be the molarity of the solution with respect to `OH^(-)`? (Assume no loss due to evaporation)

Explanation

No. of equivalents of `OH^(-)` = no. of equivalents of `Cl_(2)`
`=(10^(3))/(35.5)=28.17" "("equivalent weight of "Cl_(2)=35.5)`
Equivalent weight and formula mass of `OH^(-)` are equal.
`therefore" Number of moles of "OH^(-)=28.17`
`therefore" "[OH^(-)]=("number of moles")/("volume (in L)")=(28.17)/(20)=1.408M`