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(B) Step $1$: Calculate the moles of $NaOH$ to be neutralized.Moles of $NaOH = \frac{100 \ g}{40 \ g/mol} = 2.5 \ mol$.Step $2$: Determine the moles o

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$64$

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(B) Step $1$: Calculate the moles of $NaOH$ to be neutralized.Moles of $NaOH = \frac{100 \ g}{40 \ g/mol} = 2.5 \ mol$.Step $2$: Determine the moles of $H_2SO_4$ required.The neutralization reaction is $H_2SO_4 + 2NaOH ightarrow Na_2SO_4 + 2H_2O$.From the stoichiometry,$1 \ mol$ of $H_2SO_4$ reacts with $2 \ mol$ of $NaOH$.So,moles of $H_2SO_4$ required $= \frac{2.5}{2} = 1.25 \ mol$.Step $3$: Determine the moles of $S$ atoms required.Since $1 \ mol$ of $H_2SO_4$ contains $1 \ mol$ of $S$ atoms,$1.25 \ mol$ of $H_2SO_4$ requires $1.25 \ mol$ of $S$ atoms.Step $4$: Calculate the molecular weight of the solid.The solid contains $20\%$ by mole of $S$ atoms.Let the total moles of the solid be $n$.Moles of $S = 0.20 	imes n = 1.25 \ mol$.Therefore,$n = \frac{1.25}{0.20} = 6.25 \ mol$.Molecular weight of the solid $= \frac{	ext{Total mass}}{	ext{Total moles}} = \frac{400 \ g}{6.25 \ mol} = 64 \ g/mol$.

Find the molecular weight of a solid containing $20\%$ by mole of $S$ atoms,whose $400\ g$ provides just sufficient $S$ atoms to produce enough $H_2SO_4$ to neutralise $100\ g$ $NaOH$.

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