Question

The freezing point of a solution containing 0.2 g of acetic acid is 20.0 g of benzene is
lowered by 0.45°C, calculate :
The molar mass of acetic acid
(113.77 g mol-1)
il. van't Hoff factor
(i=0.52)
(For benxene, k, = 5.12 K kg mol-1)
What conclusion can you draw from the value of van't Hoff factor obtained.​

Answer 1

Van't Hoff Factor :

Answer :

M2 = 113.7 g/mol

Explanation :

Refer the attached picture.

Answer 2

Acetic acid associate in benzene

Explanation:

Depression in freezing point = $0.45^oC$

Mass of solute (acetic acid)  = 0.2 g

Mass of benzene (solvent) = 20.0 g = 0.02 kg    (1 kg = 1000 g)

Formula used :

$\Delta T_f=K_f\times m\\\\\Delta T_f=K_f\times\frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{Mass of water in Kg}}$

where,

$\Delta T_f$ = change in freezing point = $0.45^0C$

$K_f$ = freezing point constant = $5.12^0C/kgmole$

m = molality

Now put all the given values in this formula, we get

$0.45=5.12\times \frac{0.2g}{Mg/mol\times 0.02kg}$

M= 113.77

$i=\frac{\text {Normal Molar mass}}{\text { Observed Molar mass}}=\frac{60}{113.77}=0.52$

As i < 1 ,Thus acetic acid does associate in benzene

Learn more about depression in freezing point

https://brainly.com/question/3605961

https://brainly.in/question/13402847