A dilute solution of an ionic compound $\mathrm{A}_3 \mathrm{~B}$ has an Osmotic pressure which is 6 times that of $0.02 \mathrm{M}$ $\mathrm{MgCl}_2$. What is the molar concentration of $\mathrm{A}_3 \mathrm{~B}$ assuming that it undergoes complete dissociation in water?

The osmotic pressure $\Pi$ is given by the formula $\Pi = iCRT$, where $i$ is the van't Hoff factor, $C$ is the molar concentration, $R$ is the gas constant, and $T$ is the temperature.

For $\mathrm{MgCl}_2$, which dissociates as $\mathrm{MgCl}_2 \rightarrow \mathrm{Mg}^{2+} + 2\mathrm{Cl}^{-}$, the van't Hoff factor $i_1 = 1 + 2 = 3$. The concentration $C_1 = 0.02 \text{ M}$. Thus, $\Pi_1 = 3 \times 0.02 \times RT = 0.06 RT$.

For $\mathrm{A}_3\mathrm{B}$, which dissociates as $\mathrm{A}_3\mathrm{B} \rightarrow 3\mathrm{A}^{+} + \mathrm{B}^{3-}$, the van't Hoff factor $i_2 = 3 + 1 = 4$. Let the concentration be $C_2$. Thus, $\Pi_2 = 4 \times C_2 \times RT = 4 C_2 RT$.

Given that $\Pi_2 = 6 \times \Pi_1$, we substitute the expressions:

$4 C_2 RT = 6 \times (0.06 RT)$

$4 C_2 = 0.36$

$C_2 = \dfrac{0.36}{4} = 0.09 \text{ M}$.

Answer: 0.09 M