1.

A 0.01m aqueous solution of K_(3)[Fe(CN)_(6)] freezes ar -0.062^(@)C. What is the apparent percentage of dissociation? [K_(f) for water = 1.86]

Answer»

Solution :We have, `DeltaT_(f)=mxxK_(f)` …………(Eqn 7)
`=0.01xx1.86`
i.e., `(DeltaT_(f))_("normal")=0.0186^(@)`
and `(DeltaT_(f))_("observed")=0.062^(@)` (given)
`:.i=("observed COLLIGATIVE property")/("normal colligative property")`…………(Eqn. 9)
`=(0.062)/(0.0186)`
Now suppose x is the degree of DISSOCIATION of `K_(3)[Fe(CN)_(6)]`
Thus,
`{:("moles before DISS": :,, 1"MOLE",,0,,0),(K_(3)[Fe(CN)_(6)],,=,,3K^(+),,+[Fe(CN)_(6)]^(3-)),("moles after diss" : ,,(1-x),,3x,,x):}`
`:.i=((1-x)+3x+x)/(1)=(0.062)/(0.0186)`..........(Eqn.10)
`x=0.78`
`:.` per cent disscoiation `=78%`


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