Question

If 0.5 amp current is passed through acidified silver nitrate then in 100 minutes the mass of silver, deposite on cathode is (eq. wt. of silver nitrate + 108).

पर्याय

• 2.3523 g

• 3.3575 g

• 5.3578 g

• 6.3775 g

Answer 1

3.3575 g

Explanation:

Current (I) = 0.5 amp

Time (t) = 100 mm = 100 × 60 = 6000 s

Equivalent mass of Ag (E) = 108

w = `("Elt")/96500 - (108 xx 0.5 xx 6000)/96500` = 3.3575 g