Question:hard

At T(K), 0.004 M $Na_2SO_4$ solution is isotonic with 0.01 M glucose solution. The degree of dissociation of $Na_2SO_4$ is:

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van't Hoff factor $i = 1 + (n-1)\alpha$ for dissociation.
Updated On: Jun 6, 2026
  • 80 %
  • 50 %
  • 25 %
  • 75 %
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The Correct Option is D

Solution and Explanation

Step 1: What does isotonic mean?
Two solutions are isotonic when they push with the same osmotic pressure. Since osmotic pressure is $\pi = i C R T$ and the temperature is the same, we just need $i_1 C_1 = i_2 C_2$.

Step 2: Pick the van't Hoff factor for glucose.
Glucose does not break up in water. It stays as one particle, so its van't Hoff factor is $i = 1$.

Step 3: Write the factor for sodium sulphate.
$Na_2SO_4$ splits into $2Na^+$ and one $SO_4^{2-}$, giving $n = 3$ ions. With degree of dissociation $\alpha$, the factor is \[ i = 1 + (n-1)\alpha = 1 + 2\alpha \]

Step 4: Match the two osmotic pressures.
Glucose side: $1 \times 0.01$. Salt side: $(1 + 2\alpha) \times 0.004$. Setting them equal: \[ 0.01 = (1 + 2\alpha)(0.004) \]

Step 5: Solve for the dissociation.
Divide both sides by $0.004$: $1 + 2\alpha = 2.5$. So $2\alpha = 1.5$ and $\alpha = 0.75$.

Step 6: Turn it into a percentage.
$\alpha = 0.75$ means 75 percent of the salt has broken up.

Step 7: Conclusion.
So the answer is 75 %. \[ \boxed{75\%} \]
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