Answer :
The amount of grams of ethanol present in molar heat of vaporization of ethanol is 564.81 g.
vaporisation is the process by which a material is transformed from its liquid or solid state into its gaseous (vapour) state. Boiling is the term for the vaporisation process when circumstances permit the creation of vapour bubbles within a liquid. Sublimation is the process of directly converting a solid into a vapour.
To cause vaporisation, heat must be applied to a solid or liquid. Insufficient heat from the environment may originate from the system itself in the form of a drop in temperature. The cohesive forces that hold the atoms or molecules of a liquid or solid together must be overcome in order to separate the atoms or molecules to produce the vapour; the heat of vaporisation is a direct indicator of these forces.
The data given in the question is as follows:-
Provided heat (Q): 473.4 kJ
Molar heat of vaporization of ethanol (ΔH°vap): 38.6 kJ/mol
The formula is given as:
Q = H°vap x n
n = Q/H°vap
= 473.4/38.6
n = 12.26 mol.
The molar mass of ethanol is 46.07 g/mol:
12.26 mol x 46.07 g/mol = 564.81 g.
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