# How many moles of electron weigh one kilogram ?

$\begin{array}{1 1}(a)\;6.023\times10^{23}&(b)\;\large\frac{1}{9.108}\normalsize \times 10^{23}\\(c)\;\large\frac{6.023 \times 10^{54}}{9.108}&(d)\;\large\frac{1}{9.108\times 6.023}\normalsize \times 10^8\end{array}$

$9.108\times 10^{-31}kg$=1electron
$1kg=\large\frac{1}{9.108\times 10^{-31}}$electron
$\Rightarrow \large\frac{1}{9.108\times 10^{-31}}\times \large\frac{1}{6.023\times 10^{23}}$mole electron
$\Rightarrow \large\frac{10^8}{9.108\times 6.023}$
Hence (d) is the correct answer.