Question

Sea water is found to contain $5.85\%$ NaCl and $9.50\%$ $MgC{l}_2$ by weight of solution. calculate the normal boiling point assuming $80\%$ ionisation for NaCl and $50\%$ ionisation for $MgC{l}_2$. $\left(K_b\right(H_{2}O)=0.51kgmo{l}^{-1}K)$

• A. $T_b={101.9}^{0}C$
• B. $T_b={102.4}^{0}C$
• C. $T_b={108.5}^{0}C$
• D. $T_b={110.3}^{0}C$

Answer 1

The correct option is B $T_b={102.4}^{0}C$

$\begin{array}{c}\text{The molar masses of NaCl and}{MgCl}_2\text{are}58.5g/mol\\ \text{and}95\text{g/mol respectively.}100g\text{solution will contain}5.85g\text{NaCl}\\ \text{and}9.5gMg{Cl}_2\text{and.}\end{array}$

$100-5.85-9.5=84.65\text{g water}$

$\begin{array}{c}\text{The no. of moles of NaCl are}\\ \Rightarrow \frac{5.85}{58.5}=0.1\\ \text{The}\text{10. of moles of}{MgCl}_2\text{are}\\ \Rightarrow \frac{9.5}{95}=0.1\end{array}$

$\begin{array}{c}\text{- 80 \% ionisation of NaCl will give}\\ \begin{array}{c}\Rightarrow 0.1\times 2\times 0.8+0.1\times 0.2\\ \therefore 0.16+0.02=0.18\text{moles of particles.}\end{array}\end{array}$

$\begin{array}{c}\text{-}50\%\text{ionistation of}{MgCl}_2\text{will give}\\ \begin{array}{cc}\therefore 0.1\times 0.5\times 3+0.1\times 0.5& =\\ \Rightarrow 0.15+0.5=0.2\text{moles of particles}& \end{array}\end{array}$

$\begin{array}{c}\Rightarrow \text{Hence, the molarity of the solution}\\ \Rightarrow \frac{0.18+0.2}{0.08465}=4.489m\end{array}$

$\begin{array}{c}\text{- The elevation in the boiling point of the solution is}\\ \Delta T_b=k_b\times m\\ =0.51\times 4.489\\ ={2.3}^{\circ }C\end{array}$

$\begin{array}{c}\text{Hence, the boiling point of the solution}\\ \text{is}=100+2.3\\ ={102.3}^{\circ }C\end{array}$