pH of a 0.1 M monobasic acid is found to be 2. Hence its osmotic pressure at a given temperatue T K is 

$\begin{array}{l}\mathrm{HA}\to {\mathrm{H}}^{+}+{\mathrm{A}}^{-},\mathrm{pH}=2\\ 0.1\\ \mathrm{pH}=2 \\ \text{so }[\mathrm{H}+]=0.01\end{array}$

$$\begin{gathered} \left[{\mathrm{H}}^{+}\right]= \mathrm{C\alpha } \\ \mathrm{\alpha }=\frac{\left[{\mathrm{H}}^{+}\right]}{\mathrm{C}}=\frac{0.01}{0.1}=0.1 \\ \mathrm{i} = 1+\mathrm{\alpha }(\mathrm{n}-1) \\ \mathrm{i} = 1+0.1(2-1) \\ \mathrm{i} = 1.1 \end{gathered}$$

$\mathrm{\pi }=\mathrm{iCRT}=1.1*0.1\mathrm{RT}=0.11\mathrm{RT}$