MyQuestionIcon

1

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Interstitial Compounds

For strong el...

Question

For strong electrolytes, van't Hoff factor 'i' equals to total number of _____ (ions/atoms) in the formula unit.

Open in App

Solution

Consider the strong electrolyte $A_{x} B_{y}$ .
It completely dissociates into aqueous solution.

$A_{x} B_{y} \to x A^{+ y} + y B^{- x}$

The total number of ions in the formula unit $= x + y$
The van't Hoff factor $, i = \frac{Actual number of particles}{Number of particles for no ionization} = \frac{x + y}{1} = x + y$
Hence, for strong electrolytes, van't Hoff factor 'i' equals to total number of ions in the formula unit.

Suggest Corrections

thumbs-up

0

Similar questions

Q. Which is the correct expression of van't Hoff factor for dissociation of electrolyte?
Here,
n is the number of ions formed from one formula unit of the substance

Q. The van't Hoff factor (i) for a dilute aqueous solution of the strong electrolyte barium hydroxide is :

Q. For a non-electrolytic solution, van't Hoff factor is equal to:

For electrolytic solution, the Van't Hoff factor (i) is always equals to unity.

Q. For a solution of a non-electrolyte in water, the Van't Hoff factor is :

View More

Join BYJU'S Learning Program

similar_icon

Related Videos

lock

Interstitial Compounds

CHEMISTRY

Watch in App

explore_more_icon

Explore more

Interstitial Compounds

Standard XII Chemistry

Join BYJU'S Learning Program

CrossIcon