Question

The boiling point of benzene rises from 80.1°c to 82.4°c when 13.76 g of diphenyl is dissolved into 100g of benzene. Calculate the latent heat of vapourisation of benzene

Answer 1

Explanation:

In addition, the pi bonds in benzene are significantly less reactive than 'normal' pi bonds, either isolated or conjugated. Something about the structure of benzene makes its pi bonding arrangement especially stable. This 'something' has a name: it is called 'aromaticity

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