Question

If $3.01\times {10}^{20}$ molecules are removed from 98 mg of $H_{2}S{O}_4$, then the 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. $9.95\times {10}^{-2}mol$

Answer 1

The correct option is B $0.5\times {10}^{-3}mol$

98 mg of $H_{2}S{O}_4=6.022\times {10}^{23}\times {10}^{-3}=6.022\times {10}^{20}$
$H_{2}S{O}_4$ left = $6.022\times {10}^{20}-3.011\times {10}^{20}$
$=3.011\times {10}^{20}$
$n=\frac{N}{N_A}=\frac{3.011\times {10}^{20}}{6.022\times {10}^{23}}=0.5\times {10}^{-3}mol$