1.

The degree of ionization of 0.2 m weak acid HX is 0.3. If the K_(f) for water = 1.85. What will be the freezing point of solution ?

Answer»

`-0.360^(@)C`
`-0.206^(@)C`
`+0.480^(@)C`
`-0.480^(@)C`

Solution :`{:(,HX,hArr,H^(+),+,X^(-)),("INITIAL mole","1mole",,0,,0),("Final mole",(1-0.3),,0.3,,0.3):}`
Total mole `= (1-0.3)+0.3+0.3=1.3`
`i=(1.3)/(1)=1.3`
`Delta T_(f)=ixxmxxK_(f)`
`= 1.3xx1.85xx0.2`
`= 0.480^(@)C`
`THEREFORE` Freezing point `= 0^(@)C-0.480^(@)C`
`= -0.480^(@)C`


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