`1 g` of monobasic acid in `100 g` of water lowers the freezing point by `0.168^(@)`. If `0.2 g` of same acid requires `15.1 mL mol^(-1)` of `N//10` alkali for complete neutralization, calculate the degree of dissociation of acid. `K'_(f)` for `H_(2)O` is `1.86 K mol^(-1) kg`.

1.0 g of a monobassic acid HA in 100 g water lowers the freezing point by 0.155 K. IF 0.75 g, of same acid requires 25 mL of N/5 NaOH solution for complete neutralisation then %, degree of ionization of acid is ( K_(f) of H_(2)O = 1.86 K kg "mol"^(-1) ):

1.0 g of a monobassic acid HA in 100 g water lowers the freezing point by 0.155 K. IF 0.75 g, of same acid requires 25 mL of N/5 NaOH solution for complete neutralisation then %, degree of ionization of acid is ( K_(f) of H_(2)O = 1.86 K kg "mol"^(-1) ):

1.0 g of a monobassic acid HA in 100 g water lowers the freezing point by 0.155 K. IF 0.75 g, of same acid requires 25 mL of N/5 NaOH solution for complete neutralisation then %, degree of ionization of acid is ( K_(f) of H_(2)O = 1.86 K kg "mol"^(-1) ):

1.0 g of a weak monobasic acid HA in 100 g water lowers the freezing point by 0.155 K. If 0.45 g, of same acid requires 15 mL of N/5 NaOH solution for complete neutralistion then %, degree ofionization of acid is (K_f " of H"_(2)O="1.86 K kg mol"^(-1)) . a)0.2 b)0.25 c)0.4 d)0.5

1.5g of monobasic acid when dissolved in 150g of water lowers the freezing point by 0.165^(@)C.0.5g of the same acid when titrated, after dissolution in water, requires 37.5 mL of N//10 alkali. Calculate the degree of dissociation of the acid (K_(f) for water = 1.86^(@)C mol^(-1)) .

1.5g of monobasic acid when dissolved in 150g of water lowers the freezing point by 0.165^(@)C.0.5g of the same acid when titrated, after dissolution in water, requires 37.5 mL of N//10 alkali. Calculate the degree of dissociation of the acid (K_(f) for water = 1.86^(@)C mol^(-1)) .