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

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Q. The mass of $70\% $ $H_{2}SO_{4}$ required for neutralisation of $1$ mole of $NaOH$ is

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Solution:

$H_{2}SO_{4}$ furnishes two protons, so 1/2 mole of $H_{2}SO_{4}$ is required for complete neutralization of 1 mole of NaOH.
So, for $100\% H_{2}SO_{4}, 49\,g$ of $H_{2}SO_{4}$ is required
$\therefore $ For $70\% H_{2}SO_{4} \frac{49\times 70}{100}=34\,g$ is required