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The mass of $70 %$ (W/W) $H_{2} {SO}_{4}$ solution required for neutralization of 1 mole of $N a O H$ is?

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35 gr

70 gr

90 gr

50 gr

answer is B.

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Detailed Solution

Eq. of $H_{2} {SO}_{4} = Eq of NaoH \Rightarrow \frac{w}{49} = 1$

Wt of $H_{2} {SO}_{4} req. = 49 gr$

Wt of solution req. $= \frac{49 \times 100}{70} = 70 g r$

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