A certain liquid has a vapor pressure of 92.0 Torr at 23.0 ∘ C and 378.0 Torr at 45.0∘C.
A. Calculate the value of Δ H ∘ vap for this liquid. Δ H ∘ vap = kJ / mol.
B. Calculate the normal boiling point of this liquid. Boiling point: ∘ C.

Explanation:

(a)   The given data is as follows.

    = 92.0 torr,   = (23 + 273)K = 296 K

  = 378.0 torr,   = (45 + 273)K = 318 K

According to Clasius-Clapeyron equation,  

Putting the given values into the above formula as follows.

      1.413 =

        = 2926063.008 J/mol

or,                  = 2926.063 kJ/mol

Hence, heat of vaporization for this liquid is 2926.063 kJ/mol .

(b)   It is known that normal atmospheric pressure is 760 torr. Hence, we will calculate the normal boiling point as follows.

Hence, putting this value into Clasius-Clapeyron equation as follows.

          0.698 = 351944.07 [\frac{(318 - T)}{318T}][/tex]

               6.306 \times 10^{-6} \times T = (318 - T)

                        T = 317.998 K

Therefore, normal boiling point of this liquid is 317.998 K.

2b2t

explanation:

2b2t