Question

If $3.01 \times 10^{20}$ molecules are removed from $98 \, mg$ of $H_2SO_4$, then number of moles of $H_2SO_4$ left are

• A. $0.1 \times 10^{-3} \, mol$
• B. $0.5 \times 10^{-3} \, mol$
• C. $ 1.66 \times 10^{-3} \, mol$
• D. $9.95 \times 10^{-3} \, mol$

Answer 1

Answer: (B)

$98 \,mg$ of $H _{2} SO _{4}=6.02 \times 10^{23} \times 10^{-3}=6.02 \times 10^{20}$

Number of $H _{2} SO _{4}$ left

$=6.02 \times 10^{20}-3.01 \times 10^{20} $

$=10^{20}(6.02-3.01)=3.01 \times 10^{20} $

$\because \, n =\frac{N}{N_{A}} $

$=\frac{3.01 \times 10^{20}}{6.02 \times 10^{23}} $

$=0.5 \times 10^{-3}\, mol$