Calculate the molar enthalpy of vaporization (in kJ/mole; of a substance if 35.0 g of the substance with molar mass 16.08 p/mol, absorbs 6.411 kJ when it evaporates.

Question

Process and answer is required !

Answer 1

35.0 g x 1 mol / 16.08 g = 2.177 moles of substance (assuming molar mass is in g / mol)

q = m∆Hvap

q = heat = 6.411 kJ

m = mass = 2.177 moles

∆Hvap = enthalpy of vaporization = ?

Solving for ∆Hvap, we have ...

∆Hvap = q / m = 6.411 kJ / 2.177 moles

∆Hvap = 2.945 kJ / mol = 2.95 kJ / mol (3 sig. figs.)

Answer 2

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Step 1. Find moles of substance

Moles = Mass/Molas Mass = 35 g/ 16.08 g/mol = 2.17₆₆ mol of substance

Step 2. Calculate Molar Enthalpy

ΔH = q/n = 6.411 kJ/ 2.17₆₆ mol of substance = 2.95 kJ/mol

Notice how the sign is positive, give that the substance absorbs heat.

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