A flask contains 98 mg of H_2SO_4. If 3.01 t | vedclass

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(B) Initial weight of $H_2SO_4$ $(W)$ = $98 \ mg = 98 	imes 10^{-3} \ g$. Initial number of moles $(n_1)$ = $\frac{98 	imes 10^{-3} \ g}{98 \ g/mol}

Vedclass · Accepted Answer

$5 	imes 10^{-4}$

Vedclass · Answer

(B) Initial weight of $H_2SO_4$ $(W)$ = $98 \ mg = 98 	imes 10^{-3} \ g$. Initial number of moles $(n_1)$ = $\frac{98 	imes 10^{-3} \ g}{98 \ g/mol} = 10^{-3} \ mol$. Number of moles removed $(n_2)$ = $\frac{	ext{Number of molecules}}{N_A} = \frac{3.01 	imes 10^{20}}{6.02 	imes 10^{23}} = 0.5 	imes 10^{-3} \ mol$. Remaining moles = $n_1 - n_2 = 10^{-3} - 0.5 	imes 10^{-3} = 0.5 	imes 10^{-3} \ mol = 5 	imes 10^{-4} \ mol$.

$A$ flask contains $98 \ mg$ of $H_2SO_4$. If $3.01 \times 10^{20}$ molecules of $H_2SO_4$ are removed from the flask,the number of moles of $H_2SO_4$ remaining in the flask is $\left(N_A = 6.02 \times 10^{23}\right)$

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