`2g` of benzoic acid dissolved in `25g` of `C_(6)H_(6)`. Shows a depression in freezing point equal to `1.62K`. Molal depression constant of `C_(6)H_(6)` is `4.9K kg "mol"^(-1)`. What is the percentage association of acid if it forms double molecules in solution?

Answer

`alpha=0.992` or `99.2%`

Explanation

`because DeltaT=(1000xxk_(f)xxw)/(mxxW)
Given, w=2g,W=25g,DeltaT=1.62,K_(f)=4.9`
`therefore 1.62=(1000xx4.9xx2)/(25xxm)`
or `m_("exp")=241.98`
`because For association,
`nC_(6)H_(6)COOOHhArr(C_(6)H_(5)COOH)_(n)`
`{:(1,0),(1-alpha,alpha//n):}`
Total moles at equilibrium `=1-alpha+(alpha//n)`
`=1-alpha+(alpha//2)=1-(alpha//2)`
`n=2` (for dimer formation)
`(m_(N))/(m_("exp")=1-(alpha)/(2)` or `1-(alpha)/(2)=(122.0)/(241.98)`
`therefore alpha=0.992` or `99.2%`