Question

What is the boiling point of a solution of 9.04 g of i2 in 75.5 g of benzene, assuming the i2 is nonvolatile?

Answer 1

Answer: 374.2 K

Explanation:- Elevation in boiling point is:

`\Delta T_b=k_b* \frac{\text{mass of solute}}{\text{molar mass of solute}* \text{weight of solvent in kg}}`

`T_f` = change in boiling point

`k_b` = boiling point constant=
`2.53Kkg/mol`

m = molality

Given: mass of solute
`(I_2)`= 9.04 g

Molar mass of solute
`(I_2)`= 254 g/mol

Weight of solvent (benzene)= 75.5 g= 0.0755 kg

`\Delta T_b=2.53 Kkg/mol* (9.04g)/(254g/mol* 0.0755kg)=1.2K`

`\Delta T_b=1.2 K=T_b-T^0_b=T_b-373K`

`T_b=374.2K`

The boiling point of the solution is
`374.2K`.

Answer 2

Answer is: temperature is 81,39°C.
m(I₂) = 9,04 g.
m(C₆H₆) = 75,5 g = 0,0755 kg.
Tb(C₆H₆) = 80,2°C.
n(I₂) = m(I₂) ÷ M(I₂).
n(I₂) = 9,04 g ÷ 253,9 g/mol.
n(I₂) = 0,035 mol.
b = 0,035 mol ÷ 0,0755 kg.
b = 0,47 mol/kg.
ΔT = 0,47 mol/kg ·2,53 kg·°C/mol.
ΔT = 1,19°C.
T(solution) = 1,19°C + 80,2°C = 81,39°C.
b - molality.