Question

The elevation in boiling point of a solution of 9.43g of $MgC{{l}_{2}}$ in 1kg of water is:
a.) 0.156
b.) 0.52
c.) 0.17
d.) 0.94

Answer 1

Hint: When the solute is added to the solvent there is an increase in boiling point of solution in comparison to boiling point of pure solvent this increase in boiling point is known as elevation in boiling point. It is calculated by formula $\Delta Tb=i\times Kb\times molarity$

Complete step-by-step answer:
The correct option is A.
For calculating elevation in boiling point, we need to know van't Hoff factor, molality and Kb​,
$\Delta Tb=i\times Kb\times molarity$

Molar mass can be calculated by adding the sum of each element present in the compound.
Molar mass of $MgC{{l}_{2}}$ = 95.2g/mol
Number of moles in any solution is calculated by dividing the given mass to molecular mass.
$Number\text{ }of\text{ }moles=\text{ }\dfrac{given\text{ }moles}{molecular\text{ }mass}$
Mass of $N{{a}_{2}}S{{O}_{4}}$ in solution= 9.43g
Number of moles in $MgC{{l}_{2}}$= $\dfrac{9.43}{95.211}$
=0.09 moles
$Mass\text{ }of\text{ }water=1kg$
$Molality~(m)=~\dfrac{moles MgC{{l}_{2}}}{mass of water}=\dfrac{0.09}{1}mol/kg$

The van't Hoff factor is the dimension that is unitless; it indicates how the molecule is dissociated in solution. Since, $MgC{{l}_{2}}$ is completely dissociated in solvent therefore its van't hoff factor is 3.
The solution freezes at a temperature higher than the boiling point of solvent. This property of solutions is called elevation in boiling point.
The expression for elevation in point is shown as:
$\Delta Tb=i\times Kb\times molarity$

Where i is van't hoff factor, $\Delta Tb$ is change in boiling point ,m is molality and ${{k}_{b}}$ is ebullioscopic constant.

$\Delta Tb=3\times 0.52\times 94.3\times 19.43$​
$\Delta Tb=0.15K$
So, elevation in boiling point is 0.156K

Note: Elevation in boiling point is calculated by formula $\Delta Tb=i\times Kb\times molarity$
For which we should know van't hoff factor and molality. Molality is found by calculation the number of moles of solute and then dividing them by mass of solvent that is taken in kg.