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  4. The weight of sodium carbonate required to prepare 500 mL of a seminormal solution is

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Q. The weight of sodium carbonate required to prepare $500\, mL$ of a seminormal solution is

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Solution:

$N = \frac{w \times 1000}{eq. wt. \times \text{volume in ml.}} $

$\left( eq. wt. = \frac{106}{2} = 53 \right)$

or $w = \frac{0.5 \times 53 \times 500}{1000} = 13.25$.