The mass of $70 %$ $H_{2} S O_{4}$ required for neutralisation of $1$ mol of $N a O H$ is:

49

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98

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70

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34.3

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Solution

The correct option is C $70$ g
Equivalent of $N a O H$ $=$ Equivalent of $H_{2} S O_{4}$

Equivalent weight of $H_{2} S O_{4} = \frac{98}{2} = 49$

$\frac{W}{49} = 1$ [Equivalent $=$ mole x V.F.]

$W$ $=$ $49$ g

Total weight of $70 %$ pure sample required for neutralisation $= \frac{49 \times 100}{70} = 70$ g.

Hence, the correct option is $C$

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