Question

$36mL$ of pure water takes $100$ sec to everyone from a vessel and heater connected to as electric source which delivers $806watt.$ The $\Delta H_{vaporization}$ of $H_{2}O$ is:

• A. $40.3kJ/mol$
• B. $43.2kJ/mol$
• C. $4.03kJ/mol$
• D. none of these

Answer 1

The correct option is A $40.3kJ/mol$

Solution:- (A) $40.3KJ/mol$

As we know that,

$E=P\times t$

whereas,

$E=$ Energy consumption

$P=$ Power delivered $=806\text{watt}$

$t=$ time taken $=100\text{sec}$

$\therefore E=806\times 100=80.6KJ$

$\text{Density}=\frac{\text{mass}}{\text{volume}}$

Given volume of water $=36mL$

Density of water $=1gm/mL$

$\therefore$ Mass of $36mL$ of water $=36\times 1=36g$

Molecular weight of water $=18g$

Using, $\text{no. of moles}=\frac{\text{wt.}}{\text{mol. wt.}}$

No. of moles of water in $36g$ of water $=\frac{36}{18}=2\text{moles}$

$\therefore {\Delta H}_{vap.}=\frac{\text{Total energy consumpted}}{\text{no. of moles}}=\frac{80.6}{2}=40.3KJ$

Hence ${\Delta H}_{vapourisation}=40.3KJ/mol$.