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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

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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

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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

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