Answer to Question #187001 in General Chemistry for Dylan

Question #187001

How many grams of water can be boiled at 100oC with the application of 10,207.9 J

1
Expert's answer
2021-04-30T04:51:30-0400

Latent heat of Vapourisation = 40.8 kj / mol .


Latent heat of Vapourisation

= ( 40.8 × 1000 )j / mol. = 4800 j / mol .


Total heat = n × latent heat of Vapourisation.


n = mole = 10207.9 / 4800 mol .


n = mole = 2.13 mol.


Mass of H2O = ( 2.13 × 18 ) g = 38.34 g .


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