1.

How many atoms of hydrogen are liberated at cathode , when 965 coulombs of charge is passed through water ?

Answer»

`6.02 xx 10^(21)`
`6.02 xx 10^(23)`
`6.02 xx 10^(-19)`
`6.02 xx 10^(19)`

Solution :`H_(2)O to H_(2) + (1)/(2) O_(2) or 2 H^(+) + 2E^(-) to 2H` i.e., 96500 C produce 1 MOLE H-ATOMS = `6.02 xx 10^(23)` atoms . So 965 C will produce H-atoms = `6.02 xx 10^(21)`.


Discussion

No Comment Found

Related InterviewSolutions