# How many moles of electron weight one kilogram ?

$(a)\;6.023 \times 10^{23} \\ (b)\;\large\frac{1}{9.108} \times \normalsize 10^{31}\\(c)\; \frac{6.023}{9.108} \times 10^{54} \\(d)\;\large\frac{1}{9.108 \times 6.023} \times \normalsize 10^8$

Weight of 1 mole of electrons
$\qquad= 9.108 \times 10^{-31} \times 6.023 \times 10^{23}\;kg$
or number of moles of electron in $9.108 \times 10^{-31} \times 10^{23}\;kg$
$\qquad=1mole$
Number of moles of electrons in $1\;kg=\large\frac{1}{9.108 \times 10^{-31} \times 6.023 \times 10^{23}}=\frac{1}{9.108 \times 6.023} \times$$10^{8}\;moles$
Hence d is the correct answer.