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0.6 mL of acetic acid (CH(3)COOH) having...

`0.6 mL` of acetic acid `(CH_(3)COOH)` having density `1.06 g mL^(-1)` is dissolved in `1 L` of water. The depression in freezing point observed for this strength of acid was `0.0205^(@)C`.Calculate the Van't Hoff factor and dissociation constant of the acid. `K_(f)` for `H_(2)O=1.86 K kg ^(-1) "mol"^(-1))`

Text Solution

Verified by Experts

`Mw_(2)(CH_(3)COOH)=60 g "mol"^(-1)`,
`W_(2)(CH_(3)COOH)=v xx d`
`=0.6xx1.06`
`=0.636 g`
Molality (m)of acetic acid,
`(W_(2)xx1000)/(Mw_(2)xxW_(1))` [[1 L of `H_(2)O=1000` g],[Since `d_(H_(2)O)= 1 g mL^(-1)]]`
`=(0.636xx1000)/(60xx1000)=0.0106 "mol" kg^(-1)`
Substituting the values in the equation
`DeltaT_(f)=iK_(f)xxm`
`0.0205 K=i xx 1.86 K kg^(-1) xx 0.0106 "mol" kg^(-1)`
`:. i=1.041`
Acetic acid is a week electrolyte and dissociates in water.
Let x is the degree of dissociation of acetic acid. Thus,

Total number of moles at equiibrium
`=c(1-chi)+cx+cx=c(1+x)`
`:.i="Number of moles at equilibrium"/"Number of moles initially"=(c(1+x))/c`
`:.1.041=1+x rArr x= 0.041`
`=(0.0106mxx0.041xx0.041)/(1.00-0.041)`
`=1.86xx10^(-5)`
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