1g of a monabasic acid when dissolved in 100g of water lowers the freezing point by 0.168^(@)C. 0.2g of the same acid when dissolved and titrated required 15.1mL of N//10 alkali . Calculate the degree of dissociation of the acid. (K_(f) for water is 1.86)

1g of a monabasic acid when dissolved in 100g of water lowers the free

1.	1g of a monabasic acid when dissolved in 100g of water lowers the freezing point by 0.168^(@)C. 0.2g of the same acid when dissolved and titrated required 15.1mL of N//10 alkali . Calculate the degree of dissociation of the acid. (K_(f) for water is 1.86)
Answer» Solution :Milliequivalent of alkali (m.e.) `=(1)/(10)xx15.1=1.51` (Eqn. 1, Chapter 7) m.e. of the ACID `=1.51` (Eqn. 2 , Chapter 7) `:.` eq. of acid `=(1.51)/(1000)=0.00151` (Eqn.3, Chapter 7) Now equivalent of acid `=("weight in grams")/("equivalent weight")` `=(0.2)/("eq.wt.")` `=(0.2)/(M)` [for monobasic acid , eq. wt. =MOL. wt. (M)] `:.(0.2)/(M)=0.00151`, `M=132.45` `:.` MOLALITY of the acid (m)= `("moles of solute")/("wt. of solvent in grams")xx1000` `=(1)/(132.45)xx(1000)/(100)=0.076` We have, `(DeltaT_(f))_("normal")=K_(f).m` `=1.86xx0.076=0.141` `:.i=((DeltaT_(f))_("observed"))/((DeltaT_(f))_("normal"))`.......(Eqn 9) `=(0.168)/(0.141)=1.19` The monobasic acid (say AH) ionises as moles before diss : `{:(1"mole",,,,0,,,,0),(AH,,=,,A^(-),,+,,H^(+)):}` moles after diss : `{:((1-x),,x,,x,,(x="degree of dissociation")):}` `i=(1-x+x+x)/(1)=1.19` ..............(Eqn 10) `:.x=0.19`