$36 \mathrm{~mL}$ of pure water takes $100 \mathrm{sec}$ to evaporate from a vessel and heater connected to an electric source which delivers 806 watt. The $\Delta \mathrm{H}_{ ext {vap }}$ of $\mathrm{H}_{2} \mathrm{O}$ is :

Question

$36 \mathrm{~mL}$ of pure water takes $100 \mathrm{sec}$ to evaporate from a vessel and heater connected to an electric source which delivers 806 watt. The $\Delta \mathrm{H}_{	ext {vap }}$ of $\mathrm{H}_{2} \mathrm{O}$ is :

Aicharya · Accepted Answer

1 watt $=1 \mathrm{~J} / \mathrm{sec}$
Total heat supplied for $36 \mathrm{~mL} \mathrm{H}_{2} \mathrm{O}$
$\begin{array}{rl} & =806 \times 100 \\ & =80600 \mathrm{~J} \\ \Delta \mathrm{H}_{\mathrm{vap}} & =\frac{80600}{36} \times 18 \\ =40300 & \mathrm{~J} \text { or } 40.3 \mathrm{~kJ} / \mathrm{mol}\end{array}$

$36 \mathrm{~mL}$ of pure water takes $100 \mathrm{sec}$ to evaporate from a vessel and heater connected to…

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