Question

The molar enthalpy of vaporization of benzene at its boiling point (353K) is 29.7 KJ mol. For how long (minute) would a 11.4 Volt source need to supply a 0.5 A Current in order to vaporise 7.8 g of the, sample at its boiling point?

Answer 1

Solution:-

Molecular weight of benzene $=78g/mol$

Weight of benzene $=7.8g$

No. of moles of benzene $=\frac{7.8}{78}=0.1\text{mole}$

Given that energy provided to evaporate $1$ mole of benzene $=29.7kJ/mol$

$\therefore$ Energy provided to evaporate $0.1$ mole of benzene $\left(E\right)=29.97\times 0.1=2997J$

As we know that,

$E=V\times I\times t$

Given that,

$V=11.4\text{volt}$

$t=?$

$I=0.5A$

$2997=11.4\times 0.5\times t$

$\Rightarrow t=\frac{2997}{5.7}=525.79s$

Hence for $t=525.79s$, a $11.4$ volt source need to supply a $0.5A$ current in order to vaporise $7.8g$ of the sample at its boiling point.