Answer
(a) $1.5\times10^{-11}\,A$
(b) $4.7\times10^{7}\,ions$
Work Step by Step
(a) According to Ohm's law,
$I=\frac{V}{R}=\frac{75\times10^{-3}\,V}{5.0\times10^{9}\,\Omega}=1.5\times10^{-11}\,A$
(b) $I=\frac{q}{t}\implies q=It=1.5\times10^{-11}\,A\times0.50\,s$
$=7.5\times10^{-12}\,C$
Let the number of $Na^{+}$ ions be n.
Then, $n=\frac{q}{+e}=\frac{7.5\times10^{-12}\,C}{1.6\times10^{-19}\,C/ion}=4.7\times10^{7}\,ions$
