Question

the boiling point of benzene Rises from 80.1 degree Celsius to82.4 degree Celsius when 13.7 6 gram of diphenyl is dissolved in 100 gram of benzene calculate the latent heat of vapourisation of benzene

Answer 1

Given that,

Initial temperature = 80.1°C

final temperature = 82.4°C

Initial weight = 13.76 gram

Final weight = 100 gram

We need to calculate the latent heat of vaporization of benzene

Using formula of latent heat of vaporization

$k_{b}=\dfrac{RT^2}{1000\times l_{v}}$

$l_{v}=\dfrac{RT^2}{1000\times k_{b}}$

Where, R = gas constant

T = boiling temperature

Put the value into the formula

$l_{v}=\dfrac{2\times(353.1)^2}{1000\times2.53}$

$l_{v}=98.5\ J/kg$

Hence, The latent heat of vaporization of benzene is 98.5 J/kg.

Answer 2

ΔT = (k x w₂ x 1000)/(w₁ x M₂)

where, w₁ = 100g, w₂ = 13.76g, M₂ = 154g mol⁻¹, ΔT = (82.4−80.1)K

substitute the given values in here and you will get k = 2.574 kg mol⁻¹

Now, k = ( M₁RT₀²)/(1000 x ΔHvap)

where, M₁ = 78g mol⁻¹

substitute the given values in here and you will get ΔHvap = 31411 J mol⁻¹ = 31.411 kJ mol⁻¹