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20 July, 09:06

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.

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  1. 20 July, 10:42
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    The molar enthalpy of vaporization in kJ/mol for butane = 24.265 kJ/mol.

    Explanation:

    Firstly, we need to calculate the no. of moles of 4.00 g of liquid butane C₄H₁₀:

    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.
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