Question

If the number of electrons present in $1990mg$ of perchlorate ion is $a\times {10}^{23}$ then the value of $‘a^{\prime }$ would be? Give your answer upto two decimal only.
(Given $N_A=6.02\times {10}^{23}$, molar mass of $Cl=35.5g$)

Answer 1

$\text{Molar mass of}Cl{O}_4^{-}=35.5+16\times 4g=99.5g$
Number of moles $=\frac{1990\times {10}^{-3}}{99.5}=20\times {10}^{-3}$
$\text{number of electrons in one}Cl{O}_4^{-}=17+32+1=50$
Therefore, number of moles of electron $=20\times {10}^{-3}\times 50=1$
Hence, number of electrons =$1\times 6.02\times {10}^{23}=6.02\times {10}^{23}$
Hence, $a=6.02$