How many g of a dibasic acid (Mol. wt. = 200) | vedclass

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(A) For a dibasic acid,the equivalent weight $E$ is given by $E = \frac{M}{2} = \frac{200}{2} = 100 \ g/eq$. The formula for normality $N$ is $N = \fr

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$1$

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(A) For a dibasic acid,the equivalent weight $E$ is given by $E = \frac{M}{2} = \frac{200}{2} = 100 \ g/eq$. The formula for normality $N$ is $N = \frac{W 	imes 1000}{E 	imes V(ml)}$,where $W$ is the weight in grams and $V$ is the volume in $ml$. $A$ decinormal solution means $N = \frac{1}{10} = 0.1 \ N$. Substituting the values: $0.1 = \frac{W 	imes 1000}{100 	imes 100}$. $0.1 = \frac{W 	imes 1000}{10000} = \frac{W}{10}$. Therefore,$W = 0.1 	imes 10 = 1 \ g$.

How many $g$ of a dibasic acid $(Mol. wt. = 200)$ should be present in $100 \ ml$ of its aqueous solution to give a decinormal strength?

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