1.

A current of 96.5 A is passed for 18 min between nickel electrodes in 500 ml solution of 2 M Ni(NO_3)_2. The molarity of solution after electrolysis would be

Answer»

0.46 M
0.92 M
0.625 M
1.25 M

Solution :Moles of `NI(NO_(3))_(2)` in 500 mL of 2M `Ni(NO_(3))_(2)` is
`=(2xx500)/(1000)=1` MOL
Charge `=96.5xx18xx60=104220C`
`Ni^(2+)+2e^(-)toNi`
`2xx96500`C deposits 1 mol of `Ni(NO_(3))_(2)`
`therefore104220C` will DEPOSIT=`(104220)/(2xx96500)=0.54mol`
so, moles `Ni(NO_(3))_(2)` LEFT
`=1.0-0.54=0.46`mol
Thus, molarity of `Ni(NO_(3))_(2)` solution
`=2xx0.46=0.92mol//l`


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