1 g of a monobasic acid in 100 g of water lowers the freezing point by 0.168^(@). If 0.2 g of the same acid requires 15 mL of N/10 alkali for complete neutralisation, the degree of dissociation of the acid is (K_(f)" for water "="1.86 K kg mol"^(-1))

1 g of a monobasic acid in 100 g of water lowers the freezing point by

1.	1 g of a monobasic acid in 100 g of water lowers the freezing point by 0.168^(@). If 0.2 g of the same acid requires 15 mL of N/10 alkali for complete neutralisation, the degree of dissociation of the acid is (K_(f)" for water "="1.86 K kg mol"^(-1))
Answer» `9.8%` `19.6%` `4.9%` `1.68%` SOLUTION :`M_(2)" (mol mass of acid)"=(1000xxK_(f)xxw_(2))/(w_(1)xxDeltaT_(f))` `=(1000xx1.86xx1)/(100xx0.168)=110.71` i.e., Observed MOLECULAR mass= 110.71 15.1 mL of `(N)/(10)` alkali neutralise acid = 0.2 g `=(0.2)/(15.1)xx1000xx10=132.45g` `therefore"Eq. wt. of the acid = 132. 45"` Mol. mass of the acid = 132.45 (as it ismonobasic) This is calculated (NORMAL) mol. mass `i=("Normal mol. mass")/("Observed mol. mass")=(132.45)/(110.71)=1.196` `{:(,HA ,hArr,H^(+),+,A^(-)),("Inital",1,,0,,0),("After disso. ",1-alpha,,alpha,,alpha),(,,,,,"Total "=1+alpha):}` `therefore i=1+alpha or alpha=i-1=0.196=19.6%`