Benzene Question!! HELP NEEDED!!?

Question

Benzene is an organic solvent with the formula, C6H6. It boils at 80.1 C and melts at 5.5 C. Benzene's density is 0.88 g/ml, its heat of vaporization is 20.0 KJoules per mole, its heat of fusion is 10.0 KJoules per mole, its heat capacity is 0.50 KJoules per mole-degree C for liquid benzene, 0.25 KJoules per mole-degree C for solid benzene, and 0.125 KJoules per mole-degree C for gaseous benzene.

Calculate the following:
A.) The energy required to heat 390 gramsof benzene from 300 Celsius to 600 Celsius.
B.) The energy required to cool 3.90 micrograms of benzene from 60 Celsius to 20 Celsius.
C.) The energy removed to heat 39 kilograms of benzene from -100 Celsius to -50 Celsius.
D.) The volume occupied by 1 kilomole of liquid benzene at 25 Celsius and the volume occupied by 1 nanomole of benzene as a vapor at STP (assume an ideal gas).

Answer 1

A.) The energy required to heat 390 gramsof benzene from 300 Celsius to 600 Celsius.

moles of benzene = 390g/ 78 =5mol
since at 300-600 degC, benzene is a gas,
Cp =20kJ/mol-degC

Q= moles x Cp x (T1-T2)
= 5mol x 0.125kJ/mol-degC x (600-300)
=187.5kJ

B.) The energy required to cool 3.90 micrograms of benzene from 60 Celsius to 20 Celsius.

moles of benzene = 3.9 x 10^-6g /78 =5 x 10^-8 mol

since at 20-60 degC, benzene is a liquid,
Cp =0.50 kJ/mol-degC

Q= moles x Cp x (T1-T2)
= 5 x10^-8mol x 0.50kJ/mol-degC x (60-20)
= 10^-6kJ
= 0.001J

C.) The energy removed to heat 39 kilograms of benzene from -100 Celsius to -50 Celsius.

moles of benzene = 39000g /78 =500 mol
since at -100 to -50 degC, benzene is a solid
Cp =0.25 kJ/mol-degC
Q= moles x Cp x (T1-T2)
= 500 mol x 0.25kJ/mol-degC x (-50+100)
=6250kJ

D.) The volume occupied by 1 kilomole of liquid benzene at 25 Celsius and the volume occupied by 1 nanomole of benzene as a vapor at STP (assume an ideal gas).

since i mol of gas occupies 22.4 L,
thus 10^-9 mol of benzene vapor occupies 22.4 x 10^-9
= 2.24 x 10^-8 L =0.0224uL

For 1000mol of liquid benzene,
P=101300Pa
R=8.314J/mol.K
T= 25deg C =298K
V =nRT/P
= 1000 x 8.314 x 298/101300 =24.46m3

Answer 2

Boiling and melting points of C6H6 required.