1.

Freezing point of 0.2 M KCN solution is -0.7^(@)C. On adding 0.1 mole ofHg(CN)_(2) to one litre of the 0.2 M KCN solution, the freezing point of the solution becomes -0.53^(@)Cdue to the reaction Hg(CN)_(2)+mcN^(-)toHg(CN)_(m+2)^(m-) . What is the value of m assuming molality = molarity?

Answer»

Solution :`0.7=2xxK_(F)xx0.2=K_(f)=0.7/0.4=7/4`
`HG(CN)_(2)+mCN^(-)toHg(CN)_(m+2)^(m-)`
`0.1""0.2""0`
`0""(0.2-0.1m)""0.1`
Now, Molarity of `K^(+)=0.2M`
Molarity of `CN^(-)=(0.2-0.1m)M`
Molarity of complex `=0.1M`
`0.53=K_(f)(0.2+0.2-0.1m+0.1)=K_(f)(0.5-0.1m)`
`implies0.53=7/4(0.5-0.1m)implies2.1=3.5-0.7m`
`implies0.7m=1.4impliesm=2`


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