Solute A is a ternary electrolyte and solute B is a nonelectrolyte. If 0.1 M solution of solute B produces an osmotic pressure of 2P, then 0.05 M solution of A at the same temperature will produce an osmotic pressure equal to
(A) 1.5P
(B) 2P
(C) 3P
(D) P

Hint: Solute A gives three ions during the electrolytic dissociation. Solute B does not give away any ions. These give us the Van't Hoff factor. Substitute this along with the given details in question on the equation of osmotic pressure will give the osmotic pressure of A.

Complete step by step answer:
  -As we know, osmotic pressure is the minimum pressure that must be applied to a solution to prevent the flow of solvent molecules through a semipermeable membrane. Osmotic pressure is a colligative property and it depends on the concentration of solute particles in a solution.
 -We are given two solutes; A and B in which A is ternary electrolyte and B is a nonelectrolyte.
The equation for finding osmotic pressure can be written as follows
   $\pi =C\times R\times T\times i$ $\pi =C\times R\times T\times i$
Where $\pi $ is the osmotic pressure
            C is the molar concentration of solution
            R is the molar gas constant
            $i$ is the Van't Hoff factor

Let's consider solute A. Since it’s a ternary solute, it will dissociate three ions in the system and hence the value of $i$ is 3(${{i}_{2}}$ ) . Its concentration is given as 0.05 M(${{C}_{2}}$.).We need to find the osmotic pressure produced by the solute A(${{\pi }_{2}}$)
In the case of solute B, since it’s a non-electrolyte, its $i$ value would be 1(${{i}_{1}}$). Its concentration is given as 0.1(${{C}_{1}}$). Osmotic pressure produced by B is given as 2P (${{\pi }_{1}}$).

 As given in the question, these two take place at the same temperature. Hence by comparing both, we can write the above equation of osmotic pressure as follows
          $\dfrac{{{\pi }_{1}}}{{{C}_{1}}{{i}_{1}}}=\dfrac{{{\pi }_{2}}}{{{C}_{2}}{{i}_{2}}}$
On substituting the given values, the equation becomes
$\frac{2P}{0.1}=\dfrac{{{\pi }_{2}}}{0.05\times 3}$ $\dfrac{2P}{0.1}=\dfrac{{{\pi }_{2}}}{0.05\times 3}$
${{\pi }_{2}}=\dfrac{2P\times 0.05\times 3}{0.1}$ ${{\pi }_{2}}=\dfrac{2P\times 0.05\times 3}{0.1}$
               = 3P
So, the correct answer is “Option C”.

Note: The answer can also be found through another method.
  For A osmotic pressure, P =0.05 × 3 × R × T = 0.15RT
  For B given, 2P = 0.1 × R × T = 0.05 RT
 0.15 is 3 times 0.05.This implies that the osmotic pressure produced by A would be 3P.