Question

If $3.01\times 10^{20}$ molecules are removed from $98mg$ of $H_{2}S{O}_4$, then number of moles of $H_{2}S{O}_4$ 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)

$98mg$ of $H_{2}S{O}_4=6.02\times 10^{23}\times 10^{-3}=6.02\times 10^{20}$

Number of $H_{2}S{O}_4$ left

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

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

$\because n=\frac{N}{N_A}$

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

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