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 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.

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kobenhavn kobenhavn · Answer 2 · 2019-03-04T13:12:13+01:00

Answer: 374.2 K

Explanation:- Elevation in boiling point is:

[tex]\Delta T_b=k_b\times \frac{\text{mass of solute}}{\text{molar mass of solute}\times \text{weight of solvent in kg}}[/tex]

[tex]T_f[/tex] = change in boiling point

[tex]k_b[/tex] = boiling point constant=[tex]2.53Kkg/mol[/tex]

m = molality

Given: mass of solute [tex](I_2)[/tex]= 9.04 g

Molar mass of solute [tex](I_2)[/tex]= 254 g/mol

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

[tex]\Delta T_b=2.53 Kkg/mol\times \frac{9.04g}{254g/mol\times 0.0755kg}=1.2K[/tex]

[tex]\Delta T_b=1.2 K=T_b-T^0_b=T_b-373K[/tex]

[tex]T_b=374.2K[/tex]

The boiling point of the solution is [tex]374.2K[/tex].