Question

Calculate the boiling point of solution when 4g of $MgS{O_4}\left( {M = 120gmo{l^ - }} \right)$ was dissolved in 100g of water, assuming $MgS{O_4}$ undergoes complete ionization.
$\left[ {{K_b}\,\,{\text{for water = 0}}{\text{.52}}kgmo{l^{ - 1}}} \right]$

Answer 1

Hint- To solve this question, we will use the concept of elevation of boiling point. As we know that when a small amount of solute is added to the solvent, the boiling point of the solution increases and the elevation in the boiling point is directly proportional to the concentration of the solute in the solution. So we will begin with calculating the molality of the solute in the solution and then proceed accordingly.

Complete step by step solution:
Formula used- The elevation in boiling point is given by
$\Delta {T_b} = i \times {K_b} \times m$
Where
$\Delta {T_b}$ is the elevation in the boiling point
$i$ is the van't Hoff factor
${K_b}$ is ebullioscopic constant
m is the molality of the solute
Given
Mass of $MgS{O_4}$ = 4g
Molar mass of $MgS{O_4}$ = $120gmo{l^{ - 1}}$
Mass of water = 100g = 0.1kg
The number of moles of the $MgS{O_4}$
$moles = \dfrac{{mass\,\,of\,\,solute}}{{molar\,mass\,of\,solute}} \\
  moles = \dfrac{{4g}}{{120gmo{l^{ - 1}}}} \\
  moles = 0.0333mol \\ $
Now we will calculate the molality of the solute Which is given by
 $molality = \dfrac{{{\text{moles of MgS}}{{\text{O}}_4}}}{{{\text{ mass of water in kg}}}}$
Substituting the values of the moles of solute and mass of water
$molality = \dfrac{{{\text{moles of MgS}}{{\text{O}}_4}}}{{{\text{ mass of water in kg}}}}3 \\
  molality = \dfrac{{0.0333mol}}{{{\text{ 0}}{\text{.1kg}}}} \\
  molality = 0.333m \\ $
To calculate the van't Hoff factor is calculated as when on electrolysis the number of ions produced in the solution is equal to the van't Hoff factor
Therefore
$MgS{O_4}\xrightarrow{{}}M{g^{2 + }} + SO_4^{2 - }$
The molecule dissociate into two ions, therefore i = 2
Now we will calculate the elevation in the boiling point using the formula
$\Delta {T_b} = i \times {K_b} \times m$
Substituting the value of the above quantities
$
  \Delta {T_b} = i \times {K_b} \times m \\
  \Delta {T_b} = 2 \times 0.52 \times 0.333 \\
  \Delta {T_b} = 0.3466K \\
$
The normal boiling point of the water at room temperature is 100 degree Celsius or 373.15K
Due to addition of the solute the increase in the boiling point of water is
$
   = 373.15 + 0.3466 \\
   = 373.5K \\ $

The new boiling point of the solution is 373.5K

Note- In order to solve these types of questions, you need to have a good concept of solute, solvent and solution. Also the boiling point is the temperature at which vapour pressure is equal to the atmospheric pressure or surrounding pressure. With the addition of solute the boiling point of the solution increases because the greater amount of heat is needed to boil the solution. In the above question we calculated the concentration of the solute in the solution and then using the formula of elevation calculated the increase in temperature.