Answer to Question #187001 in General Chemistry for Dylan

Question

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

Answer 1

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 .