KVPY SA Exam 2015 Question Paper

Question : 74 of 80

Marks: +1, -0

The amount of Ag (atomic mass = 108) deposited at the cathode when a current of 0.5 amp is passed through a solution of

A gNO_3

for 1 hour is closest to

Solution:

W=\Eit/96500

={108×0.5× 3600}/96500

= 2 gm

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