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 $E = \frac{M}{n} = \frac{200}{2} = 100 \ g/eq$.The formula for normality $N$ is $N = \fra

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

List-$I$	List-$II$
$a$. Sodium laurylsulphate	$i$. Toilet soap
$b$. Cetyltrimethyl ammonium chloride	$ii$. Non-ionic detergent
$c$. Sodium stearate	$iii$. Anionic detergent
$d$. Polyethyleneglycyl stearate	$iv$. Cationic detergent

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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