20 ml of a decinormal solution of NaOH neutralizes 25ml of a solution of dibasic acid containing 3g of the acid per 500ml. The molecular weight of the acid is

By Ostwald's dilution law :

$N_1{V}_1=N_2{V}_2$

Where $N_1$and $N_2$ are the normalities of the first and second solutions respectively 

$V_1$ is the volume of solution used to neutralise the $V_2$ volume of second solution.

So, on putting the given values in the equation we get :

$$\begin{gathered} N_1{V}_1=N_2{V}_2 \\ \frac{1}{10}\times 20=(n-\text{ factor }\times \text{ molarity })\times 25 \end{gathered}$$

Now the n-factor for the dibasic acid is 2, therefore:

$\frac{1}{10}=\times 20=2\times \frac{\left(\frac{\text{ weight }}{\text{ molecularweight }}\right)}{\text{ Volume }\left(l\right)}\times 25$

Or, ${}^2=2\times \frac{\frac{3}{\text{ molecellar weight }}}{\frac{\text{ sout }}{\text{ 1000 }}}\times 25$
$\mathrm{Or},\frac{6}{\text{ molecular weight }}\times 25=1$

Or, Molecular weight $=150 \mathrm{g}$

Hence, Option A is the right choice