If 2.4088 times 10^{21} molecules are removed | vedclass

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(D) The molar mass of $H_2SO_4 = 98 \, g/mol$.The given mass of $H_2SO_4 = 392 \, mg = 392 	imes 10^{-3} \, g = 0.392 \, g$.The number of moles of $H

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(D) The molar mass of $H_2SO_4 = 98 \, g/mol$.The given mass of $H_2SO_4 = 392 \, mg = 392 	imes 10^{-3} \, g = 0.392 \, g$.The number of moles of $H_2SO_4 = \frac{0.392 \, g}{98 \, g/mol} = 0.004 \, mol = 4 	imes 10^{-3} \, mol$.The total number of molecules initially present $= 4 	imes 10^{-3} 	imes 6.022 	imes 10^{23} = 2.4088 	imes 10^{21}$ molecules.Since $2.4088 	imes 10^{21}$ molecules are removed,the remaining molecules $= 2.4088 	imes 10^{21} - 2.4088 	imes 10^{21} = 0$.

If $2.4088 \times 10^{21}$ molecules are removed from $392 \, mg$ of $H_2SO_4$ in a container,the number of $H_2SO_4$ molecules remaining in the container will be ........

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