(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.

(a) When 19.5 g of F-CH_(2)-COOH (Molar mass = 78" g mol"^(-1)) is dis

1.	(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.
Answer» Solution :(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.