Question

Which one of the following is an isotonic pair of solution ?

• A. 0.01 M Bacl₂ and 0.001 M CaCl₂
• B. 0.01 M Bacl₂ and 0.0015 MNaCl
• C. 0.001 M Al₂(SO₄)₃and 0.01 M BaCl₂
• D. 0.001 M CaCl₂ and 0.001 M Al₂(S0₄)₃

Answer 1

The Correct Option is B

To determine which pair of solutions is isotonic, we need to compare their osmotic pressures. The osmotic pressure depends on the concentration of solute particles, which can be determined using the formula: \[ \text{Osmotic pressure} = i \times M \] where:
\( i \) is the van't Hoff factor (the number of ions produced per formula unit of the solute).
\( M \) is the molarity of the solution. Now, let's calculate the osmotic pressure for each solution:
1. For BaCl\(_2\) (0.01 M): BaCl\(_2\) dissociates into Ba\(^{2+}\) and 2Cl\(^-\), so \( i = 3 \).
Osmotic pressure = \( 3 \times 0.01 = 0.03 \). 2. For CaCl\(_2\) (0.001 M):
CaCl\(_2\) dissociates into Ca\(^{2+}\) and 2Cl\(^-\), so \( i = 3 \).
Osmotic pressure = \( 3 \times 0.001 = 0.003 \). 3. For NaCl (0.015 M):
NaCl dissociates into Na\(^+\) and Cl\(^-\), so \( i = 2 \).
Osmotic pressure = \( 2 \times 0.015 = 0.03 \). 4. For Al\(_2\)(SO\(_4\))\(_3\) (0.001 M):
Al\(_2\)(SO\(_4\))\(_3\) dissociates into 2Al\(^{3+}\) and 3SO\(_4^{2-}\), so \( i = 5 \).
Osmotic pressure = \( 5 \times 0.001 = 0.005 \). Now, comparing the osmotic pressures:
Osmotic pressure of 0.01 M BaCl\(_2\) = 0.03.
Osmotic pressure of 0.015 M NaCl = 0.03.
Osmotic pressures of other pairs are not equal.
Therefore, the solutions of BaCl\(_2\) (0.01 M) and NaCl (0.015 M) are isotonic.

The correct option is (B) : 0.01 M Bacl₂ and 0.0015 MNaCl