Question

Number of atoms of iron present in $100$ g ${Fe}_2{O}_3$ having $20\%$ purity is:

• A. $0.20$ $N_A$
• B. $0.25$ $N_A$
• C. $0.50$ $N_A$
• D. $0.30$ $N_A$

Answer 1

The correct option is B $0.25$ $N_A$

$100$ g of ${Fe}_2{O}_3=\frac{100\times 20}{100}gpure{Fe}_2{O}_3$ $=\frac{20}{160}moles{Fe}_2{O}_3$ (molar mass of ${Fe}_2{O}_3=160$)
$=\frac{20}{160}\times 2moles$ $Fe$
$=0.25\times 6.023\times {10}^{23}atoms$ of $Fe$
$=0.25N_{A}atoms$ of $Fe$