Question

Calculate the boiling point elevation of 0.100 kg of water containing 0.010 mol of NaCl, 0.020 mol of $\mathbf{N}{\mathbf{a}}_2\mathbf{S}{\mathbf{O}}_4$, and 0.030 mol of $\mathbf{M}\mathbf{g}\mathbf{C}{\mathbf{l}}_2$, assuming complete dissociation of these electrolytes.

Short answer

The boiling point elevation of 0.100 kg of water containing 0.010 mol of NaCl, 0.020 mol of $\mathbf{N}{\mathbf{a}}_2\mathbf{S}{\mathbf{O}}_4$, and 0.030 mol $\mathbf{M}\mathbf{g}\mathbf{C}{\mathbf{l}}_2$of are 0.0521, 0.1042, 1.563.

Step-by-step solution

Boiling Point

Thetemperature at which liquid vapour pressure equals atmospheric pressure is referred to as boiling point.

 Step 2: Molality of the Compounds

Mass ofWater=0.100kg

Number of Moles of NaCl= 0.010mole

Number of Moles of$\mathbf{N}{\mathbf{a}}_2\mathbf{S}{\mathbf{O}}_4$ = 0.020mole

Number of Moles of$\mathbf{M}\mathbf{g}\mathbf{C}{\mathbf{l}}_2$ = 0.030mole

Mass of Solvent (Benzene) = 75.5g = 0.0755kg

$\begin{array}{rl}\mathbf{M}\mathbf{o}\mathbf{l}\mathbf{a}\mathbf{l}\mathbf{i}\mathbf{t}\mathbf{y}& ={\displaystyle \frac{\mathbf{N}\mathbf{u}\mathbf{m}\mathbf{b}\mathbf{e}\mathbf{r}\mathbf{o}\mathbf{f}\mathbf{m}\mathbf{o}\mathbf{l}\mathbf{e}\mathbf{s}}{\mathbf{M}\mathbf{a}\mathbf{s}\mathbf{s}\mathbf{o}\mathbf{f}\mathbf{S}\mathbf{o}\mathbf{l}\mathbf{v}\mathbf{e}\mathbf{n}\mathbf{t}\mathbf{(}\mathbf{k}\mathbf{g}\mathbf{)}}}\\ \mathbf{M}\mathbf{o}\mathbf{l}\mathbf{a}\mathbf{l}\mathbf{i}\mathbf{t}\mathbf{y}\mathbf{o}\mathbf{f}\mathbf{N}\mathbf{a}\mathbf{C}\mathbf{l}& ={\displaystyle \frac{\mathbf{0}\mathbf{.010}}{\mathbf{0}\mathbf{.100}}}\\ \mathbf{M}\mathbf{o}\mathbf{l}\mathbf{a}\mathbf{l}\mathbf{i}\mathbf{t}\mathbf{y}\mathbf{o}\mathbf{f}\mathbf{N}\mathbf{a}\mathbf{C}\mathbf{l}& =0\mathbf{.1}\mathbf{m}\end{array}$

$\begin{array}{rl}\mathbf{M}\mathbf{o}\mathbf{l}\mathbf{a}\mathbf{l}\mathbf{i}\mathbf{t}\mathbf{y}\mathbf{o}\mathbf{f}\mathbf{N}{\mathbf{a}}_2\mathbf{S}{\mathbf{O}}_4& ={\displaystyle \frac{\mathbf{0}\mathbf{.020}}{\mathbf{0}\mathbf{.100}}}\\ \mathbf{M}\mathbf{o}\mathbf{l}\mathbf{a}\mathbf{l}\mathbf{i}\mathbf{t}\mathbf{y}\mathbf{o}\mathbf{f}\mathbf{N}{\mathbf{a}}_2\mathbf{S}{\mathbf{O}}_4& =\mathbf{0}\mathbf{.2}\mathbf{m}\end{array}$

$\begin{array}{rl}\mathbf{M}\mathbf{o}\mathbf{l}\mathbf{a}\mathbf{l}\mathbf{i}\mathbf{t}\mathbf{y}\mathbf{o}\mathbf{f}\mathbf{M}\mathbf{g}\mathbf{C}{\mathbf{l}}_2& ={\displaystyle \frac{\mathbf{0}\mathbf{.030}}{\mathbf{0}\mathbf{.100}}}\\ \mathbf{M}\mathbf{o}\mathbf{l}\mathbf{a}\mathbf{l}\mathbf{i}\mathbf{t}\mathbf{y}\mathbf{o}\mathbf{f}\mathbf{M}\mathbf{g}\mathbf{C}{\mathbf{l}}_2& =\mathbf{0}\mathbf{.3}\mathbf{m}\end{array}$

Boiling Point of Compounds 

$\mathbf{\Delta }\mathbf{T}\mathbf{b}\mathbf{=}\mathbf{K}\mathbf{b}\mathbf{.}\mathbf{m}$

$\begin{array}{r}\mathbf{F}\mathbf{o}\mathbf{r}\mathbf{N}\mathbf{a}\mathbf{C}\mathbf{l}\\ \mathbf{\Delta }\mathbf{T}\mathbf{b}& =\mathbf{0}\mathbf{.521}\times \mathbf{0}\mathbf{.1}\\ \mathbf{\Delta }\mathbf{T}\mathbf{b}& =\mathbf{0}\mathbf{.521}\end{array}$

$\begin{array}{r}\mathbf{F}\mathbf{o}\mathbf{r}\mathbf{N}{\mathbf{a}}_2\mathbf{S}{\mathbf{O}}_4\\ \mathbf{\Delta }\mathbf{T}\mathbf{b}& =0\mathbf{.521}\mathbf{0}\times \mathbf{2}\\ \mathbf{\Delta }\mathbf{b}& =0\mathbf{.1042}\end{array}$

$\begin{array}{r}\mathbf{F}\mathbf{o}\mathbf{r}\mathbf{M}\mathbf{g}\mathbf{C}{\mathbf{l}}_2\\ \mathbf{\Delta }\mathbf{T}\mathbf{b}& =0\mathbf{.521}\times \mathbf{0}\mathbf{.3}\\ \mathbf{\Delta }\mathbf{T}\mathbf{b}& =1\mathbf{.563}\end{array}$