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  4. The mass of 70% H2SO4 by mass required for neutralisation of 1 mol of NaOH is

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Q. The mass of 70% $H_2SO_4$ by mass required for neutralisation of 1 mol of NaOH is

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Solution:

$2NaOH + H_{2}SO_{4} \to Na_{2}SO_{4}+H_{2}O$
Pure $H_{2}SO_{4}$ required for $1$ mole of $NaOH$
$=\frac{1}{2}$ mol $=49\,g$
$70\%$ $H_{2}SO_{4}$ required for $1$ mole of $NaOH$
$=\frac{49\times100}{70}$
$=70\,g$