Question

Determine the number of moles present in 0.55 mg of electrons:

• A. 1 mole
• B. 2 moles
• C. 1.5 moles
• D. 0.5 moles

Answer 1

The correct option is A 1 mole

Mass of $e^{-}=9.1\times {10}^{-31}kg$

So $9.1\times {10}^{-31}\times {10}^6$ mg contains $1e^{-}$

or $1$mg contains $\frac{1}{9.1\times {10}^{-31}\times {10}^6}{e}^{-}$

so $0.55$ mg contains $\frac{1}{9.1\times {10}^{-25}}\times 0.55e^{-}$

$=6.043\times {10}^{23}{e}^{-}$

Now simply apply avogadro's law

$6.023\times {10}^{23}{e}^{-}$ make $1$ mole

so $1$ mole $\Rightarrow$ Ans