Answer to Question #132996 in General Chemistry for Vanessa

Question

How much heat, in J, is required to convert 122.5 g of liquid benzaldehyde at 115.0 °C to vapor (gas) benzaldehyde at 207 °C? (Hint: 3 steps are need to solve this problem)


Molar mass of benzaldehyde = 106.12 g/mol

Boiling point of benzaldehyde = 179 °C

Specific heat of liquid benzaldehyde = 1.62 J/g °C

Specific heat of vapor (gas) benzaldehyde = 1.38 J/g °C

Enthalpy of fusion of benzaldehyde, ∆Hvap = 42.5kJ/mol

Answer 1

Q= mc∆T = 122.5 x 1.62 x 92 = 18 257,4 J = 18.2574 kJ