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Solutions question

2023 · 11 Apr · Shift 1 · Q17
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Solutions question

2023 · 11 Apr · Shift 1 · Q17

JEE MainChemistrySolutionsNumerical+4 / −1
0.004 M K 2_22​ SO 4_44​ solution is isotonic with 0.01 M glucose solution. Percentage dissociation of K 2_22​ SO 4_44​ is ‾\underline{\hspace{2cm}}​ (Nearest integer)
Numerical answer
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Correct answer: 75

  1. Use isotonic condition

For isotonic solutions at the same temperature:

π=iCRT\pi = i C R Tπ=iCRT

So,

i1C1=i2C2i_1 C_1 = i_2 C_2i1​C1​=i2​C2​

Here:

  • For glucose, it is a non-electrolyte, so i=1i=1i=1
  • For K2SO4K_2SO_4K2​SO4​, let van't Hoff factor be iii

Thus,

i×0.004=1×0.01i \times 0.004 = 1 \times 0.01i×0.004=1×0.01

i=0.010.004=2.5i = \frac{0.01}{0.004} = 2.5i=0.0040.01​=2.5

  1. Relate van't Hoff factor to degree of dissociation

K2SO4K_2SO_4K2​SO4​ dissociates as:

K2SO4→2K++SO42−K_2SO_4 \rightarrow 2K^+ + SO_4^{2-}K2​SO4​→2K++SO42−​

So, 1 formula unit gives 3 particles.

If degree of dissociation is α\alphaα, then

i=1+(3−1)α=1+2αi = 1 + (3-1)\alpha = 1 + 2\alphai=1+(3−1)α=1+2α

Given i=2.5i=2.5i=2.5,

2.5=1+2α2.5 = 1 + 2\alpha2.5=1+2α

2α=1.52\alpha = 1.52α=1.5

α=0.75\alpha = 0.75α=0.75

  1. Convert to percentage dissociation

% dissociation=0.75×100=75%\%\text{ dissociation} = 0.75 \times 100 = 75\%% dissociation=0.75×100=75%

  1. Nearest integer

75\boxed{75}75​

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