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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»


Solution :`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`
`becauseFor 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%`


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