If 3.01 times 10^{20} molecules are removed f | vedclass

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(B) The molar mass of $H_2SO_4$ is $98 \ g/mol$.Initial amount of $H_2SO_4 = 98 \ mg = 98 	imes 10^{-3} \ g = 0.1 	imes 10^{-2} \ mol = 1 	imes 10^

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$0.5 	imes 10^{-3} \ mol$

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(B) The molar mass of $H_2SO_4$ is $98 \ g/mol$.Initial amount of $H_2SO_4 = 98 \ mg = 98 	imes 10^{-3} \ g = 0.1 	imes 10^{-2} \ mol = 1 	imes 10^{-3} \ mol$.Number of molecules in $1 	imes 10^{-3} \ mol = 1 	imes 10^{-3} 	imes 6.022 	imes 10^{23} = 6.022 	imes 10^{20}$ molecules.Number of molecules removed $= 3.01 	imes 10^{20}$.Number of molecules left $= 6.022 	imes 10^{20} - 3.01 	imes 10^{20} = 3.012 	imes 10^{20}$ molecules.Number of moles left $= \frac{3.012 	imes 10^{20}}{6.022 	imes 10^{23}} \approx 0.5 	imes 10^{-3} \ mol$.

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

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