Question

the normal boiling point of toluene is 110.6 degree Celcius. what is its vapor pressure at this temperature

Answer 1

Use Clausius-Clapeyron-equation
d(lnp) / dT = ΔHv/(R·T²)
integrated from
ln(p(T)) = - ΔHv/(R·T) + C

If know the vapor pressure at a temperature
let say p(T₀) = p₀
you can calculate vapor at any other temperature from the equation:
p = p₀ · e^{ (ΔHv/R)·(1/T₀ - 1/T) }

At the normal boiling point vapor pressure equals atmospheric pressure. So.
T₀=110.6°C= 383.75K
p₀=1atm

Hence vapor pressure at T=87.1°C=360.25K is
p = 1atm · e^{ (39200J/mol/8.314472J/molK) · (1/383.75K - 1/360.25K) }
= 0.457atm
= 46.35kPa