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Chemistry Question on Solutions

The degree of dissociation (α)(\alpha) of a weak electrolyte AxByA_xB_y is related to van?t Hoff factor (i)(i) by the expression

A

α=i−1x+y−1\alpha=\frac{i-1}{x+y-1}

B

α=x+y−1i−1\alpha=\frac{x+y-1}{i-1}

C

α=x+y+1i−1\alpha=\frac{x+y+1}{i-1}

D

α=i−1x+y+1\alpha=\frac{i-1}{x+y+1}

Answer

α=i−1x+y−1\alpha=\frac{i-1}{x+y-1}

Explanation

Solution

AxBy→xAy++yBx−A_xB_y \rightarrow xA^{y+} + yB^{x-}
α=i−1n−1\alpha = \frac{i-1}{n-1}
where nn is the number of ions produced.
α=i−1(x+y−1)\alpha = \frac{i -1}{(x+y-1)}