JEE MAIN - Chemistry (2011 - No. 7)
The degree of dissociation ($$\alpha$$ ) of a weak electrolyte, AxBy is related to van’t Hoff factor (i) by the expression :
$$\alpha = {{i - 1} \over {x + y + 1}}$$
$$\alpha = {{x + y - 1} \over {i - 1}}$$
$$\alpha = {{x + y + 1} \over {i - 1}}$$
$$\alpha = {{i - 1} \over {(x + y - 1)}}$$
คำอธิบาย
$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{A_x}\,{B_y}\,\,\rightleftharpoons\,\,{}_x{A^{y + }} + {}_y{B^{x - }}$$
$$t = 0\,\,\,\,\,\,\,\,\,1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0$$
$${t_{eq}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1 - \alpha \,\,\,\,\,\,\,\,\,x\alpha \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,y\alpha $$
Total no. of moles $$(i)$$ $$ = 1 - \alpha + x\alpha + y\alpha $$
$$i - 1 = x\alpha + y\alpha - \alpha $$
$$ = \alpha \left( {x + y - 1} \right)$$
$$\therefore$$ $$\,\,\,\,\,\,\,\alpha = {{i - 1} \over {\left( {x + y - 1} \right)}}$$
$$t = 0\,\,\,\,\,\,\,\,\,1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0$$
$${t_{eq}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1 - \alpha \,\,\,\,\,\,\,\,\,x\alpha \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,y\alpha $$
Total no. of moles $$(i)$$ $$ = 1 - \alpha + x\alpha + y\alpha $$
$$i - 1 = x\alpha + y\alpha - \alpha $$
$$ = \alpha \left( {x + y - 1} \right)$$
$$\therefore$$ $$\,\,\,\,\,\,\,\alpha = {{i - 1} \over {\left( {x + y - 1} \right)}}$$