An experiment produces evidence that the evaporation of 4.00 g of liquid butane c4h10 (l) requires a gain in enthalpy of 1.67 kj. find the molar enthalpy of vaporization in kj/mol for butane from this evidence.

The molar enthalpy of vaporization in kJ/mol for butane = 24.265 kJ/mol.

Explanation:

n = mass/molar mass = (4.0 g)/(58.12 g/mol) = 0.0688 mol.

∴ 0.0688 mol of butane requires a gain in enthalpy of 1.67 kJ to be evaporized.

Know using cross multiplication:

0.0688 mol of butane to be vaporized requires → 1.67 kJ.

1.0 mol of butane to be vaporized requires → ??? kJ.

∴ 1.0 mol of butane to be vaporized requires = (1.0 mol)(1.67 kJ)/(0.0688 mol) = 24.265 kJ.

∴ The molar enthalpy of vaporization in kJ/mol for butane = 24.265 kJ/mol.

the answer is 0.8m.

by: nicholas