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Q)

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

$(a)\;0.2g\qquad(b)\;0.4g\qquad(c)\;0.6g\qquad(d)\;None\;of\;these$

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A)
$H_2SO_4+2NaOH\rightarrow Na_2SO_4+H_2O$
$\therefore$ millimoles of $H_2SO_4=\large\frac{1}{2}$$\times$ millimoles of NaOH
So, millimoles of NaOH =$2\times M\times V = 2\times 0.5\times 10 = 10$
Or $\large\frac{w}{m_{NaOH}}$$\times 1000=10$
$\Rightarrow \large\frac{w}{40}$$\times 1000=10$
$\Rightarrow w=0.4g$
Hence (b) is the correct answer.
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