Question

How much heat must be added to vaporize 245 g of butane if its heat of vaporization is 21.3 kJ/mol? | ΑΣφ ? Q= KJ You have already submitted this answer. Enter a new answer. No credit lost. Try again. Submit Previous Answers Request Answer Part F What volume does this much butane occupy at 760 torr and 37 °C? Η ΑΣΦ ? V= Submit Request Answer

Answer 1

E.

Molar mass of butane, C4H10 = 58 g / mol

Heat of vapourization of butane = 21.3 kJ / mol

So,

Heat of vaporization of 58 g. of butane = 21.3 kJ

Then,

Heat of vaporization of 245 g. of butane = 245 x 21.3 / 58 = 90.0 kJ

F.

Moles of butane, n = Mass / Molar mass = 245 / 58 = 4.22 mol

Pressure, P = 760 torr = 1 atm

Temperature, T = 37 + 273 = 310 K

R = 0.0821 L.atm / K.mol

Volume, V = ?
Ideal gas equation,

P V = n R T

1 x V = 4.22 x 0.0821 x 310

V = 107. L