Question

A solute gets 50% dimerised and 50% ionized into two ions. The van't Hoff factor of solute is​

Answer 1

Answer:

The van't Hoff factor, i, of the solute measured is $1.25$.

Explanation:

Given,

The dimerisation of solute, β = $50\%$ = $\frac{50}{100}$ = $0.5$.

The dissociation of solute, α = $50\%$ = $\frac{50}{100}$ = $0.5$.

The total ions formed, n = $2$

The van't Hoff factor of the given solute, i =?

As we know,

• The van't Hoff factor of solute can be calculated by the equation given below:
• i = $\frac{(1-\alpha +n\alpha ) + (1-\beta +n\frac{\beta }{2} )}{n}$

After putting the given values in the equation, we get:

• i = $\frac{(1-0.5 +2\times 0.5 ) + (1-0.5 +2\times\frac{0.5 }{2} )}{2}$
• i = $\frac{(0.5 +1) + (0.5 +0.5)}{2}$
• i = $\frac{1.5+1}{2}$
• i = $\frac{2.5}{2}$
• i = $1.25$

Hence, the van't Hoff factor of solute, i = $1.25$.