Question

Sea water is found to contain $5.85\% \;NaCl$and $9.50\% {\text{ }}MgC{l_2}$ by
weight of solution. Calculate its boiling point assuming $80\% $ for $NaCl$ and $50\% $ ionisation of $MgC{l_2}$${K_b}({H_2}O) = 0.51kg$.
(A) ${T_b} = {101.9^ \circ }C$
(B) ${T_b} = {102.3^ \circ }C$
(C) ${T_b} = {108.5^ \circ }C$
(D) ${T_b} = {110.3^ \circ }C$

Answer 1

Hint: in the given question we are asked to find the temperature or the boiling point of the solution which is basically the product of molality and elevation constant of a solvent and we know that molality of a solution is the ratio of moles of solute to the weight of solvent in kilograms.

Complete step by step solution:
First here we need to calculate the number of moles of sodium chloride as we are given with its mass. Let us assume $100g$of sea water,$mass\;of\;NaCl = 5.85g$, so we can calculate the number of moles of $NaCl$
$
  moles = \dfrac{{given\;mass}}{{molecular\;mass}} \\
   \Rightarrow moles = \dfrac{{5.85}}{{58.5}} \\
   \Rightarrow moles = 0.1mole \\
 $
Similarly we can find the moles of $MgC{l_2}$, given mass of $MgC{l_2}$$ = 9.5g$and we also know the molecular mass of $MgC{l_2}$$ = 95g$ so the number of moles will be:
$
  moles = \dfrac{{given\;mass}}{{molecular\;mass}} \\
   \Rightarrow moles = \dfrac{{9.5}}{{95}} \\
   \Rightarrow moles = 0.1mole \\
 $
Now, we know that $NaCl$ will dissociate into sodium ions and chloride ions, so the initial moles and moles at equilibrium will be given as:
$NaCl \to \;N{a^ + } + C{l^ - }$

\[0.1\]	$0$	$0$
\[0.1 - x\]	$0$	$0$

$80\% $ ionisation of sodium chloride will give the total number of moles which will be= $(0.1 - x + x + x)$ and $0.1 - x = $$0.1 - \dfrac{{0.1 \times 80}}{{100}} = 0.2$, $x = \dfrac{{0.1 \times 80}}{{100}}$$ = 0.8$, so the total moles of $NaCl = 0.18$
Similarly, we know that $MgC{l_2}$ will dissociate into magnesium and chloride ions, so initial moles and moles at equilibrium will be:
$MgC{l_2} \to M{g^{2 + }} + C{l^ - }$

\[0.1\]	$0$	$0$
\[0.1 - x\]	$0$	$0$

$50\% $ ionisation will give the total number of moles of $MgC{l_2}$ which will be $ = (0.1 - x + x + x)$, after solving we get
$0.1 - x$$ = 0.1 - \dfrac{{0.1 \times 50}}{{100}} = 0.05$,$x = \dfrac{{0.1 \times 50}}{{100}} = 0.10$
and for chlorine it will be $x = 2 \times \dfrac{{0.1 \times 50}}{{100}} = 0.1$, thus total number of moles will be: $MgC{l_2} = 0.20$.
Now, after dissociation in sea water, the total number of both the compounds will be: $0.81 + 0.20 = 0.38$. From these moles we can calculate the molality of the solution:
$
  molality = \dfrac{{mass\;of\;solute}}{{mass\;of\;solvent}} \\
   \Rightarrow molality = \dfrac{{0.38}}{{100 - 5.85 - 9.5}} \\
   \Rightarrow molality = 4.4\;mol\;k{g^{ - 1}} \\
 $
Now, using the formula for elevation in boiling point of the solution we can calculate $\Delta {T_b}$ and we are given with ${K_b} = 0.51$ so we will get,
$
  \Delta {T_b} = {K_b} \times m \\
   \Rightarrow \Delta {T_b} = 0.51 \times 4.4 \\
   \Rightarrow \Delta {T_b} = {2.3^ \circ }C \\
 $
Now we can finally calculate the boiling point of the solution using this temperature, we will get:
$
  {T^s} = {T^ \circ } + \Delta {T_b} \\
   \Rightarrow {T^s} = 100 + 2.3 \\
   \Rightarrow {T^s} = {102.3^ \circ }C \\
 $
Therefore, the correct answer is (B).

Note: Remember that the boiling point is that temperature at which vapour pressure of a liquid is equivalent to the pressure surrounding that liquid. The normal boiling point of water is ${100^ \circ }C$, at this temperature the vapour pressure of water is equivalent to $1atm$.