Question

A current of 0.5 ampere when passed through AgNO3 solution for 193 sec deposited 0.108 g of silver. The equivalent weight of Ag is

Answer 1

Answer:  The equivalent weight of Ag is 108 grams

Explanation:

$Q=I\times t$

where Q= quantity of electricity in coloumbs

I = current in amperes = 0.5 A

t= time in seconds =$193s$

$Q=0.5A\times 193s=96.5C$

$AgNO_3\rightarrow Ag^++NO_3^-$

$Ag^{+}+e^{-1}\rightarrow Ag$

96.5 Coloumb of electricity deposits 0.108 g of Ag

96500 C of electricity deposits =$\frac{0.108}{96.5}\times 96500=108g$ of Ag.