Question

QUESTION 3 How much heat is required to convert 243.7 g of liquid benzene (Mm - 78.11 g/mol) at 75.1 °C to gaseous benzene at 115.1°C? The following information may be useful. b.p = 80.1 °C Cm liquid benzene) - 1.74g ** | Vap 30.2 kg/ Cm (gaseous benzene) -0.469/g 101965 102,0 940 36.8

Answer 1

Ans 3 : 102.0

Heat required to raise the temperature of benzene to 80.1^oC :

= mass x specific heat x change in temperature

putting values :

= 243.7 g x 1.74 J/g^oC x (80.1^oC - 75.1^oC)

= 2120.2 J

= 2.12 KJ

Heat required to vaporise benzene :

=mol x Hvap

mol benzene = mass /molar mass

= 243.7 g / 78.11 g/mol

= 3.12 mol

putting values :

= 3.12 mol x 30.72 KJ/mol

= 95.8 KJ

Heat required to raise the temperature to 115.1^oC

= 243.7 g x 0.469 J/g^oC x (115.1^oC - 80.1^oC)

= 4000 J

= 4 KJ

total heat = 2.12 KJ + 95.8 KJ + 4 KJ

= 102.0 KJ