Question

Cyclohexane has a freezing point of 6.50 ∘C and a Kf of 20.0 ∘C/m. What is the freezing point of a solution made by dissolving 0.771 g of biphenyl (C12H10) in 25.0 g of cyclohexane?

Answer 1

Answer:
`2.49^0C`

Step-by-step explanation:

Depression in freezing point is:

`T_f^0-T_f=i* k_f* (w_2* 1000)/(M_2* w_1)`

where,

`T_f` = freezing point of solution = ?

`T^o_f` = freezing point of solvent (cyclohexane) =
`6.50^oC`

`k_f` = freezing point constant of solvent (cyclohexane) =
`20.0^oC/m`

m = molality

i = Van't Hoff factor = 1 (for non-electrolyte)

`w_2` = mass of solute (biphenyl) = 0.771 g

`w_1` = mass of solvent (cyclohexane) = 25.0 g

`M_2` = molar mass of solute (biphenyl) =

Now put all the given values in the above formula, we get:

`(6.50-T_f)^oC=1* (20.0^oC/m)* ((0.771g)* 1000)/(154* (25.0g))`

`(6.50-T_f)^oC=4.01`

`T_f=2.49^0C`

Therefore, the freezing point of a solution made by dissolving 0.771 g of biphenyl in 25.0 g of cyclohexane is
`2.49^0C`