Solution has $8.35 g$ $I_{2}$ in $71.0 g$ of toluene. Assuming $I_{2}$ is nonvolatile. Toluene, $T b = {110.63}^{0} C$ and $K b = {3.40}^{0} C K g / m o l$ What is the boiling point of solution (in °C) ?

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Solution

.
$Δ T_{b} = i K_{b} m$

Where,
$i$ is the van't Hoff factor, i.e. the effective number of dissociated particles per undissociated solute particle.

$K_{b} = {3.40}^{\circ} C . K g / m o l$ is the boiling point elevation constant of toluene.

m is the molality, mol solute/kg solvent.

So, let's find the molality of a toluene iodine solution then,

$8.35 g I_{2} \times \frac{1 m o l}{253.808 g I_{2}} = 0.03290$ moles

$71.0 g$ toluene = $0.0710 k g$ toluene

Therefore, the molality is:

$m = \frac{0.03290 m o l e s o l u t e}{0.0710 k g s o l v e n t} = 0.4634 m o l / k g$

$i = 1.$

The change in boiling point is:

$Δ T_{b} = (1) ({3.40}^{\circ} C k g / m o l) (0.4634 m o l / k g)$
$= {1.58}^{\circ} C$

As a result, the new boiling point is:
$T_{b} = T_{b}^{*} + Δ T_{b}$

$= {110.63}^{\circ} C + {1.58}^{\circ} C = {112.21}^{\circ} C$

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