Verified Solution • 2025–26 Session moderate

QJEEM11Chc292: JEE Main 2011 Chemistry - Class 12 JEEM11Chc292 Solution: The degree of dissociation ( ) of a weak electrolyte, A x B y…

The degree of dissociation ( α\alpha ) of a weak electrolyte, AxBy is related to van’t Hoff factor (i) by the expression :

Multiple Choice Options

  • A α=i−1x+y+1\alpha = {{i - 1} \over {x + y + 1}}
  • B α=x+y−1i−1\alpha = {{x + y - 1} \over {i - 1}}
  • C α=x+y+1i−1\alpha = {{x + y + 1} \over {i - 1}}
  • D α=i−1(x+y−1)\alpha = {{i - 1} \over {(x + y - 1)}}

Step-by-Step Verified Solution

               Ax By  ⇌  xAy++yBx−\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{A_x}\,{B_y}\,\,\rightleftharpoons\,\,{}_x{A^{y + }} + {}_y{B^{x - }}

t=0         1                     0                      0t = 0\,\,\,\,\,\,\,\,\,1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0

teq                1−α         xα                 yα{t_{eq}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1 - \alpha \,\,\,\,\,\,\,\,\,x\alpha \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,y\alpha

Total no. of moles (i)(i) =1−α+xα+yα= 1 - \alpha + x\alpha + y\alpha

i−1=xα+yα−αi - 1 = x\alpha + y\alpha - \alpha

=α(x+y−1)= \alpha \left( {x + y - 1} \right)

∴\therefore        α=i−1(x+y−1)\,\,\,\,\,\,\,\alpha = {{i - 1} \over {\left( {x + y - 1} \right)}}

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