How many coulombs of electricity are consumed when a

 100 mA current is passed through a solution of $A{\mathrm{gNO}}_3$ 

 for half an hour during an electrolysis experiment? 

 $1\mathrm{mA}={10}^{-3}\text{ ampere }$

$\therefore$  $100\mathrm{mA}\equiv 100\times {10}^{-3}\mathrm{A}$

$$\begin{gathered} \mathrm{Time}=1/2\mathrm{hr} \\ =1/2\times 60\times 60=1800 \sec \end{gathered}$$ 

Quantity of electricity to be passed (Q) = Current (in amperes) x time (in seconds)

 $\begin{array}{r}\mathrm{Q}=100\times {10}^{-3}\mathrm{A}\times 1800 \mathrm{s}=180 \mathrm{As}\\ =180 \mathrm{C}\end{array}$