100 , mL of a mixture of NaOH and Na_2SO_4 is | vedclass

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(B) The reaction between $NaOH$ and $H_2SO_4$ is: $2NaOH + H_2SO_4 ightarrow Na_2SO_4 + 2H_2O$. Milliequivalents $(meq)$ of $H_2SO_4 = 	ext{Molarit

Vedclass · Accepted Answer

$0.4$

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(B) The reaction between $NaOH$ and $H_2SO_4$ is: $2NaOH + H_2SO_4 ightarrow Na_2SO_4 + 2H_2O$. Milliequivalents $(meq)$ of $H_2SO_4 = 	ext{Molarity} 	imes 	ext{n-factor} 	imes 	ext{Volume (in } mL) = 0.5 	imes 2 	imes 10 = 10 \, meq$. Since $meq$ of $NaOH = meq$ of $H_2SO_4$,we have $meq$ of $NaOH = 10$. $meq = \frac{	ext{mass in } g}{	ext{Equivalent weight}} 	imes 1000$. Equivalent weight of $NaOH = 40$. $10 = \frac{w}{40} 	imes 1000$. $w = \frac{10 	imes 40}{1000} = 0.4 \, g$.

$100 \, mL$ of a mixture of $NaOH$ and $Na_2SO_4$ is neutralized by $10 \, mL$ of $0.5 \, M$ $H_2SO_4$. Hence,the mass of $NaOH$ in $100 \, mL$ solution is ........... $g$.

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