:a certain liquid has a vapor pressure of 6.91 mmhg at 0 °c. if this liquid has a normal boiling point of 105 °c, what is the liquid's heat of vaporization in kj/mol?

liquid's heat of vaporization = 38.4 kJ/mol

Explanation:

given data

vapor pressure P1 = 6.91 mmHg

at temperature = 0 °C = 273.15 K

boiling temperature = 105 °C

solution

for vapor pressure and temperature we get here

P2 = 760.0 mmHg

T2 = 68.73°C = 378.15 K

we use here the Clausius-Clapeyron Equation that is

ln =     .................1

put here value

In

solve it we get

x = 38445 J/mol

liquid's heat of vaporization = 38.4 kJ/mol

i look it up online and i'll tell you

i would say b

explanation:

but i think i would need more information than that