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 `Delta H_("vaporization") of H_(2)O` is: 40.3 kJ/mol, 43,2 kJ/mol, 4.03 kJ/mol, 4.32 kJ/mol

Options

A. 40.3 kJ/mol
B. 43,2 kJ/mol
C. 4.03 kJ/mol
D. 4.32 kJ/mol

Answer

A

Explanation

1watt = 1J/sec, Total heat supplied for 36 mL `H _(2)O`
`= 806 xx 100 = 80600 J, Delta H _(vap) = ( 80600)/(36) xx 18 = 40300 J or 40.3 kJ//mol`