Question
$$36\,mL$$ of pure water takes 100 sec to evaporate from a vessel and heater connected to an electric source which delivers $$806\,watt.$$ The $$\Delta {H_{vap}}$$ of $${H_2}O$$ is :
A.
$$40.3\,kJ/mol$$
B.
$$43.2\,kJ/mol$$
C.
$$4.03\,kJ/mol$$
D.
$${\text{None of these}}$$
Answer :
$$40.3\,kJ/mol$$
Solution :
$$\eqalign{
& 1\,watt = 1\,J/\sec \cr
& {\text{Total heat supplied for}}\,\,36\,mL\,\,{H_2}O \cr
& = 806 \times 100 \cr
& = 80600\,J \cr
& \Delta {H_{vap}} = \frac{{80600}}{{36}} \times 18 \cr
& = 40300\,J\,\,{\text{or}}\,\,40.3\,kJ/mol \cr} $$