Calculate the molarity of a solution containing 5g of NaOH dissolved in the product of a `H_(2)-O_(2)` fuel cell operated at 1A curent for 595.1 hours. (Assume 1F=96500C/mol of electrons and molecular weight of NaOH as `40g mol^(-1)`)

Options

A. 0.05M
B. 0.025M
C. 0.1M
D. 0.63M

Answer

B

Explanation

Total charge produced by cell =`1 A xx (5951 xx 3600)s`
=2142360C
`because 96500C=1` mol
`therefore 2142360C= (2142360)/(96500)`
=22.20 mol of `e^(-)`
Now, in `H_(2).O_(2)` fuel cell following reaction occurs,
At anode `2H_(2)(g) + 4O^(-) H(aq) rarr 4H_(2)O (l) + 4e^(-)`
At cathode `O_(2)(g) + 2H_(2)O (l) + 4e^(-) rarr 4OH^(-)(aq)`
Overall reaction `2H_(2)(g) + O_(2)(g) rarr 2H_(2)O(l)`
Thus, from above reaction it is clear that 2mol of `e^(-) -=1` mol of `H_(2)O`
`because 22` mol of `e^(-) -=11` mol of `H_(2)O`
`=(11 xx 18)g " of " H_(2)O` [`because` NO. of moles `=("Weight")/("Molecular weight")`]
`=198 g or mL " of " H_(2)O`
Now, number of mol of NaOH `=(5)/(40)=0.125` mol
We know that, Molarity =`("No. of moles of solute")/("Volume of solution (in L)") = (0.125)/(198)xx 1000= 0.63M`