Home
Class 12
CHEMISTRY
Depression of freezing point of 0.01 mol...

Depression of freezing point of `0.01` molal aq. `CH_(3)COOH` solution is `0.02046^(@)` . 1 molal urea solution freezes at `-1.86^(@)`C . Assuming molarity equal to molarity , pH of `CH_(3)COOH` solution is :

A

2

B

3

C

`3.2`

D

`4.2`

Text Solution

Verified by Experts

The correct Answer is:
B
Promotional Banner

Topper's Solved these Questions

  • LIQUID SOLUTIONS

    GRB PUBLICATION|Exercise 33|7 Videos
  • LIQUID SOLUTIONS

    GRB PUBLICATION|Exercise 36|6 Videos
  • LIQUID SOLUTIONS

    GRB PUBLICATION|Exercise 30|7 Videos
  • IONIC EQUILIBRIUM

    GRB PUBLICATION|Exercise All Questions|526 Videos
  • METALLUGY

    GRB PUBLICATION|Exercise Subjective type|1 Videos

Similar Questions

Explore conceptually related problems

The depression in freezing point of 0.01 m aqueous CH_(3)CooH solution is 0.02046^(@) , 1 m urea solution freezes at -1.86^(@)C . Assuming molality equal to molarity, pH of CH_(3)COOH solution is

Depression in freezing point of 0.01 molal aqueous HCOOH solution is 0.02046^(@)C . 1 molal aqueous urea solution freezes at -1.86^@ C . Assuming molality equal to molarity , calculate the pH of HCOOH solution.

depression in freezing point of 0.01 molal aqueous CH_(3)COOH solution is 0.02046^(@)C . Assuming molality equal to molarity. pH of CH_(3)COOH solution is (K_(f)=1.86K Kg "mol"^(-1)) _______________________.

If the freezing point of 0.1 M HA(aq) solution is -0.2046^(@)C then pH of solution is ( If K_(f) water =1.86mol^(-1)kg^(-1))

Depression in freezing point of 0.1 molal solution of HF is -0.201^(@)C . Calculate percentage degree of dissociation of HF. (K_(f)=1.86 K kg mol^(-1)) .

The freezing point of one molal NaCl solution assuming NaCl to be 100% dissociated in water is (molal depression constant= 1.86)