Question

If a 0.890 m aqueous solution freezes at − 2.40 ∘ C, what is the van't Hoff factor, i , of the solute?

Answer 1

Answer: Van't Hoff factor, i , of the solute is 1.45

Step-by-step explanation:

Depression in freezing point is given by:

`\Delta T_f=i* K_f* m`

`\Delta T_f=T_f^0-T_f=(0-(-2.40)^0C=2.40^0C` = Depression in freezing point

i= vant hoff factor = ?

`K_f` = freezing point constant =
`1.86^0C/m`

m= molality =0.890 m

`2.40^0C=i* 1.86^0C/m* 0.890`

`i=1.45`

Thus the van't Hoff factor, i , of the solute is 1.45