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28/09/2021 8:39 pm
Topic starter
36 mL of pure water takes 100 sec to evaporate from a vessel and heater connected to an electric source which delivers 806 watt. The ΔHvap of H2O is
(a) 40.3 kJ/mol
(b) 43.2 kJ/mol
(c) 4.03 kJ/mol
(d) None of these
1 Answer
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28/09/2021 8:41 pm
Correct answer: (a) 40.3 kJ/mol
Explanation:
1 watt = 1 J/sec
Total heat supplied for 36 mL H2O
= 806 × 100
= 80600 J
ΔHvap = \(\frac{80600}{36}\) x 18
= 40300 J or 40.3 kJ/mol