(a) When 19.5 g of `F-CH_(2)-COOH` (Molar mass = `78" g mol"^(-1)`) is dissolved in 500 g of water, the depression in freezing point is observed to be `1^(@)C`. Calculate the degree of dissociation of `F-CH_(2)-COOH`.
[Given : `K_(f)` for water = 1.86 K kg `mol^(-1)`]
(b) Give reasons :
(i) 0.1 M KCl has higher boiling point than 0.1 M Glucose.
(ii) Meat is preserved for a longer time by salting.

Explanation

(a) Number of moles of `F-CH_(2)-COOH=(19.5)/(78)=0.25` moles
Molality of the solution `=(0.25)/(500)xx1000=0.5" mol kg"^(-1)`
Using the following equation
`Delta T_(f)=iK_(f)m`, where I is van.t Hoff factor
Substituting the values in the above equation, we have
`1K=ixx1.86" K kg mol"^(-1)xx0.5" mol kg"^(-1)`
or `" "i=(1)/(1.86xx0.5)=(1)/(0.93)=1.075" "...(i)`
`I-"CII"_(2)-COOH to F-underset(x)(CH_(2)COO^(-))+underset(x)(H^(+))`
where x is degree of dissociation.
Total number of moles `=1-x+x+x=1+x`
`i=(1+x)/(1)" "...(ii)`
From (i) and (ii), we have
`1+x=1.075` or x = 0.075
Degree of dissociation = 0.075 or 7.5%
(b) (i) KCl dissociates to give `K^(+)` and `Cl^(-)`. As the concentration of particles in 0.1 M KCl is more than that of 0.1 M glucose, boiling point of KCl is higher.
(ii) Whater is taken out of bacteria into the solution of sodium chloride as a result of osmosis. Therefore bacteria do not survive.