Home
Class 12
CHEMISTRY
1.5g of monobasic acid when dissolved in...

`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))`.

Text Solution

Verified by Experts

The correct Answer is:
`0.196`
Promotional Banner

Topper's Solved these Questions

  • LIQUID SOLUTION

    FIITJEE|Exercise ASSIGNMENT PROBLEMS (OBJECTIVE) level-I|55 Videos

  • LIQUID SOLUTION

    FIITJEE|Exercise ASSIGNMENT PROBLEMS (OBJECTIVE) level-II|30 Videos

  • LIQUID SOLUTION

    FIITJEE|Exercise ASSIGNMENT PROBLEMS (SUBJECTIVE) level-I (NUMERICAL BASED PROBLEMS)|8 Videos

  • IONIC EQUILIBRIUM

    FIITJEE|Exercise SINGLE INTEGER ANSWER QUESTIONS|4 Videos

  • NUCLEIC ACID AND VITAMIN

    FIITJEE|Exercise ASSIGNMENT PROBLEMS (OBJECTIVE)|15 Videos

Similar Questions

Explore conceptually related problems

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 .

A solution is prepared by dissolving 1.5g of a monoacidic base into 1.5 kg of water at 300K , which showed a depression in freezing point by 0.165^(@)C . When 0.496g of the same base titrated, after dissolution, required 40 mL of semimolar H_(2)SO_(4) solution. If K_(f) of water is 1.86 K kg mol^(-1) , then select the correct statements (s) out of the following( assuming molarity = molarity):

1 g of a monobasic acid dissolved in 200 g of water lowers the freezing point by 0.186^@ C .On the other hand when 1 g of the same acid is dissolved in water so as to make the solution 200 mL , this solution requires 125 mL of 0.1 N NaOH for complete neutralization. If alpha for the acid is x xx10 . Find x:

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) ):

5 g of a substance when dissolved in 50 g water lowers the freeezing by 1.2^(@)C . Calculate molecular wt. of the substance if molal depression constant of water is 1.86^(@)C K kg mole^(-1) .

6 g of a substance is dissolved in 100 g of water depresses the freezing point by 0.93^(@)C . The molecular mass of the substance will be: ( K_(f) for water = 1.86^(@)C /molal)