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Using the heat of vaporization in calculations
Calculate the mass of water (in g) that can be vaporized at its boiling point with 106 kj of heat. (the dhvaporization at boiling point is 40.7 kj/mol).

Respuesta :

bharathparasad577

The mass of water(in g) that can be vaporized at its boiling point(100°C) with 106 kJ of heat is "46.879g".

According to question:

ΔHvap=40.7kJ/mol

Now,

To calculate the enthalpy change of a reaction we use the formula:

ΔHreaction=Heat/Number of moles

∵Heat absorbed=106kJ

& ΔHreaction=Enthalpy change of the reaction=40.7kJ/mol

∴The number of moles evaporated by 106 kJ heat will be:

106×1÷40.7=2.6044 mol

Now,Number of moles=Given mass/Molar mass

∵Molar mass of water=18

& Number of moles of water=2.6044

Thus,Mass of water=Number of moles×Molar Mass of water

                              =2.6044×18

                               =46.879g

∴Mass of water(in g) that can be vaporized at its boiling point=46.879g

LEARN MORE ABOUT LATENT HEAT OF VAPORIZATION HERE:

https://brainly.com/question/13372553

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