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Question: The osmotic pressure of an ionic compound \[\text{XY}\] in water is four times that of a solution of...

The osmotic pressure of an ionic compound XY\text{XY} in water is four times that of a solution of 0⋅010\cdot 01M BaCl2\text{BaC}{{\text{l}}_{2}} in water. Assuming complete dissociation of the given ionic compounds in water, the concentration of XY\text{XY} (in molL−1{{\text{L}}^{-1}}) in solution is:
(A) 6×10−26\times {{10}^{-2}}
(B) 4×10−44\times {{10}^{-4}}
(C) 16×10−416\times {{10}^{-4}}
(D) 4×10−24\times {{10}^{-2}}

Explanation

Solution

The pressure that would have to be applied to a solution to stop the passage through a semipermeable membrane of solvent molecules from the pure solvent. Osmotic pressure for dilute solutions of non-electrolytes can be calculated by modifying the ideal gas equation:

& \Rightarrow \Pi V=nRT \\\ & \Rightarrow \Pi =\dfrac{n}{V}RT \\\ \end{aligned}$$ Now, $$\dfrac{n}{V}=M$$ where $$M$$ is the molar concentration of dissolved species. Thus, the equation becomes $$\Rightarrow \Pi =MRT$$ Here, $$\Pi =$$ Osmotic Pressure $$R=$$ Ideal gas constant and has value $$0\cdot 08206\text{ L}$$atm $$\text{mo}{{\text{l}}^{-1}}{{\text{K}}^{-1}}$$ $$T=$$ Temperature **Formula used:** $$\Rightarrow \Pi =icRT$$ Here, $$ic$$ is the molar concentration of the solute i in solution. **Complete Step by step solution:** Given that, osmotic pressure of an ionic compound $$\text{XY}$$ in the water is four times that of a solution of $$\text{0}\cdot \text{01 M BaC}{{\text{l}}_{2}}$$ in water. Therefore, $$\Rightarrow \Pi \text{ XY = 4}\Pi \text{ BaC}{{\text{l}}_{2}}$$ Since, both are ionic compounds, $$\Rightarrow \text{XY}\to {{\text{X}}^{+}}+{{\text{Y}}^{-}}\Rightarrow i=2$$ $$\Rightarrow \text{BaC}{{\text{l}}_{2}}\to \text{B}{{\text{a}}^{2+}}+2\text{C}{{\text{l}}^{-}}\Rightarrow i=3$$ Now, applying the formula of osmotic pressure $$\Rightarrow \Pi =icRT$$ Assuming, the temperature $$T$$ is constant. $\begin{aligned} & \Rightarrow 2\times \left[ \text{XY} \right]=4\times 3\times 0.01 \\\ & \Rightarrow \left[ \text{XY} \right]=0.06\text{ M} \\\ & \Rightarrow \left[ \text{XY} \right]=6\times {{10}^{-2}}\text{M} \\\ \end{aligned}$ **Hence, the concentration of $\left[ \text{XY} \right]$ is $\text{6}\times \text{1}{{\text{0}}^{-2}}\text{mol }{{\text{L}}^{-1}}$.** **Additional information:** Ideal gas law: It simply states that under constant pressure and temperature and pressure, the volume of gas solely depends upon the number of moles of its molecule, not the type of the gas. PV=nRT Where, P= Pressure V= Volume n= Number of moles of molecules R= Gas constant T= Temperature **Note:** The Van’t Hoff theory describes that substance in dilute solution obey the ideal gas laws. Thus, the conceptual knowledge about various topics like ideal gas law, molarity, osmotic pressure, etc. is required by the students.