A certain mass of a substance when dissolved in `100 g C_(6)H_(6)` lowers the freezing point by `1.28^(circ)C`. The same mass of solute dissolved in `100 g` of water lowers of the freezing point by `1.40^(circ)C`. If the substance has normal molecular weight in benzene and is completely dissocited in water, into how many ions does it dissocite in water ? `K_(f)` for `H_(2)O` and `C _(6)H_(6)` are `1.86` and `5.12 K mol^(-1) kg` respectively.

Answer

3

Explanation

`Delta T_(f)=(1000xxK'_(f)xxw)/(m xx W)`
For the solution in benzene using the data given
`1.28=(1000xx5.12xx w)/(m_(N)xx100)` (`m_(N)` = normal mol. mass) ……(i)
For the solution in water in which solute dissociates
`1.40=(1000xx1.86xx w)/(m_(esp.)xx100)` .......(ii)
Dividing eq. (ii) by (i).
`i=(m_(N))/(m_(exp.))=(1.40)/(1.28)xx(5.12)/(1.86)=3.01~~3.0`
Now, suppose that formula for solute is
`{:(A_(x)B_(y),hArr,xA^(+),+,yB^(-)),(1,,0,,0),((1-alpha),,x alpha,,y alpha):}`
`therefore i=1-alpha+x alpha+y alpha`
`because` i=3 and `alpha = 1` (Given that `alpha = 1`)
`therefore` No. of ions given (x+y)=3