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Solutions question

2011 · Shift 0 · Q2
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Solutions question

2011 · Shift 0 · Q2

JEE MainChemistrySolutionsMCQ+4 / −1
The degree of dissociation (α\alphaα ) of a weak electrolyte, AxBy is related to van’t Hoff factor (i) by the expression :
  1. A
    α=i−1x+y+1\alpha = {{i - 1} \over {x + y + 1}}α=x+y+1i−1​
  2. B
    α=x+y−1i−1\alpha = {{x + y - 1} \over {i - 1}}α=i−1x+y−1​
  3. C
    α=x+y+1i−1\alpha = {{x + y + 1} \over {i - 1}}α=i−1x+y+1​
  4. D
    α=i−1(x+y−1)\alpha = {{i - 1} \over {(x + y - 1)}}α=(x+y−1)i−1​
View written solutionFree

Correct answer: D

  1. Write the dissociation of the weak electrolyte

A weak electrolyte AxByA_xB_yAx​By​ dissociates as:

AxBy⇌xA⋯+yB⋯A_xB_y \rightleftharpoons xA^{\cdots} + yB^{\cdots}Ax​By​⇌xA⋯+yB⋯

So, from 1 formula unit, the total number of particles after complete dissociation would be:

x+yx+yx+y

  1. Assume degree of dissociation is α\alphaα
  • Fraction undissociated =1−α= 1-\alpha=1−α
  • Fraction dissociated =α= \alpha=α

Hence, total particles in solution become:

(1−α)(1)+α(x+y)(1-\alpha)(1) + \alpha(x+y)(1−α)(1)+α(x+y)

=1−α+α(x+y)= 1-\alpha + \alpha(x+y)=1−α+α(x+y)

=1+α(x+y−1)= 1 + \alpha(x+y-1)=1+α(x+y−1)

  1. Use definition of van't Hoff factor

Van’t Hoff factor is:

i=actual number of particles in solutioninitial number of formula unitsi = \frac{\text{actual number of particles in solution}}{\text{initial number of formula units}}i=initial number of formula unitsactual number of particles in solution​

Since initially there is 1 formula unit,

i=1+α(x+y−1)i = 1 + \alpha(x+y-1)i=1+α(x+y−1)

  1. Solve for α\alphaα

i−1=α(x+y−1)i-1 = \alpha(x+y-1)i−1=α(x+y−1)

Therefore,

α=i−1x+y−1\alpha = \frac{i-1}{x+y-1}α=x+y−1i−1​

  1. Match with the options

This corresponds to:

α=i−1x+y−1\boxed{\alpha = \frac{i-1}{x+y-1}}α=x+y−1i−1​​

So, the correct option is D.

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