Question

X is an electrolyte with a concentration of 0.04M whose formula is of the type X\(_2\)(A) Y is a non-electrolyte solution with a concentration of 0.2M and has an osmotic pressure equal to P\(_2\) at room temperature. What is the relationship between the Osmotic pressure \(\pi\) of X and P\(_2\)?

• A. \(\pi = 0.6 P_2\)
• B. \(\pi = 0.12 P_2\)
• C. \(\pi = 0.04 P_2\)
• D. \(\pi = 0.8 P_2\)

Hint:

The osmotic pressure of an electrolyte solution depends on the van’t Hoff factor. For a dissociating electrolyte, multiply the molarity by the number of ions forme(D)

Answer 1

The Correct Option is A

To find the relationship between the osmotic pressures of X and Y, we use the formula for osmotic pressure: \[ \pi = i \cdot M \cdot R \cdot T \] Where: - \(\pi\) is the osmotic pressure, - \(i\) is the van’t Hoff factor (which is 2 for \(X_2A\), as it dissociates into two ions: \(X_2A \rightarrow 2X + A\)), - \(M\) is the molarity of the solution, - \(R\) is the gas constant, - \(T\) is the temperature in Kelvin. For the electrolyte \(X_2A\), the van't Hoff factor \(i = 2\). For the non-electrolyte solution Y, \(i = 1\), since it does not dissociate. Now, let’s calculate the osmotic pressures for both solutions: - For \(X_2A\) (electrolyte): \[ \pi_X = 2 \cdot 0.04 \cdot R \cdot T = 0.08 \cdot R \cdot T \] - For Y (non-electrolyte): \[ \pi_Y = 0.2 \cdot R \cdot T \] We are given that the osmotic pressure of Y is \(P_2\), so \(\pi_Y = P_2\). Now, comparing \(\pi_X\) and \(\pi_Y\): \[ \frac{\pi_X}{\pi_Y} = \frac{0.08 \cdot R \cdot T}{0.2 \cdot R \cdot T} = 0.4 \] Thus, \(\pi_X = 0.6 \cdot P_2\).