At what temperature does benzene boil when the external pressure is 410 torr ?

The temperature at which benzene boils at a given pressure of 410 torr can be deduced from the Clausius Clapeyron equation:

ln(P2/P1) = -ΔHvap/R [1/T2-1/T1] (1)

where

ΔHvap = heat of vaporization of benzene

P1 = atmospheric pressure

T1 = normal boiling point

R = gas constant

These values are essentially standard values which can be obtained from any thermochemical data base

For benzene: ΔHvap = 30.72 kJ/mol and T1 = 80 C = 80 +273 = 353 K

R = 0.008314 kJ/K.mol

P1 = 760 torr

P2 = 410 torr

Substituting these values in equation (1) we get:

ln(410/760) = -30.72/0.008314 [1/T2-1/353]

T2 = 337 K

or, T2 = 337 - 273 = 64 C

c is having 4 valence electrons and is bonded to three hydrogens and one sulphur atom, therefore, the number of bond pair is equal to four. as all the valence electrons are used in forming the bonds so, there is no lone pair left on carbon.

s is having 6 valence electrons and is bonded to one other sulfur atom and one carbon atom, therefore, the number of bond pair is equal to 2. as only two valence electrons are used in forming the bonds so, four electrons are left. therefore, two lone pairs are present on s.