If a 0.004 M solution of Na2SO4 is isotonic with a 0.010 M solution of glucose at same temperature. Then what is the apparent degree of dissociation of Na2SO4 is

Question

If a 0.004 M solution of Na2SO4 is isotonic with a 0.010 M solution of glucose at same temperature. Then what is the apparent degree of dissociation of Na2SO4 is (A) 25% (B) 50% (C) 75% (D) 85%

Tardigrade Admin · Accepted Answer

Isotonic solutions have equal osmotic pressures.

(0.004 - x) N a_{2} S O_{4} ⇌ 2 x 2 N a^{+} + x S O_{4}^{2 -}

π_{1} = π_{2} i_{1} \times C_{1} \times RT = i_{2} \times C_{2} \times RT i_{1} \times C_{1} = i_{2} \times C_{2} i_{1} \times 0.004 = 1 \times 0.01 i_{1} = 2.5 i_{1} = 1 + 2 αα = \frac{2.5 - 1}{2} = 0.75% α = 0.75 \times 100 = 75%

or
Since both the solution are isotonic

0.004 + 2 x = 0.01

x = 0.003

Percent dissociation of

N a_{2} S O_{4} = \frac{3 \times 1 0 ^{- 3}}{0.004} \times 100 = 75%

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Q. If a 0.004 M solution of Na2​SO4​ is isotonic with a 0.010 M solution of glucose at same temperature. Then what is the apparent degree of dissociation of Na2​SO4​ is

25%

50%

75%

85%

Q. If a 0.004 M solution of $N a_{2} S O_{4}$ is isotonic with a 0.010 M solution of glucose at same temperature. Then what is the apparent degree of dissociation of $N a_{2} S O_{4}$ is