Question

The elevation in boiling point, when 13.44 g of freshly prepared $\text{ CuC}{{\text{l}}_{\text{2}}}\text{ }$ are added to one kilogram of water is: [Some useful data,$\text{ }{{\text{K}}_{b}}\text{ }=\text{ }0.52\text{ kg Kmo}{{\text{l}}^{-1}}\text{ }$ . The molecular weight of$\text{ CuC}{{\text{l}}_{\text{2}}}\text{ = }134.4\text{ gm }$]
A) $\text{ 0}\text{.05 }$
B) $\text{ 0}\text{.1 }$
C) $\text{ 0}\text{.16 }$
D) $\text{ 0}\text{.21 }$

Answer 1

The elevation of the boiling point $\text{ }\\!\\!\Delta\\!\\!\text{ }{{\text{T}}_{\text{b}}}$ is a colligative property. It depends on the amount of solute. If the solute undergoes the association or dissociation in the solution then the difference in the boiling point is stated as:
$\text{ }\\!\\!\Delta\\!\\!\text{ }{{\text{T}}_{\text{b}}}\text{ = i m }{{\text{K}}_{\text{b}}}\text{ }$

Complete step by step answer:
The boiling point, $\text{ }\\!\\!\Delta\\!\\!\text{ }{{\text{T}}_{\text{b}}}\text{ }$ of a liquid, is the temperature at which the vapour pressure is equal to the atmospheric pressure. When a non-volatile solute is added to a liquid, the vapour pressure of the liquid is decreased. Hence, it must be heated to higher temperatures so that its vapour pressure becomes equal to that of the atmospheric pressure. This means that the addition of a non-volatile solute to a liquid raises its boiling point.
The inorganic salts may associate or dissociate into the solution. Therefore, the elevation is the boiling point is related to the van’t Hoff factor, molality of the solution, and ebullioscopic constant as follows,
$\text{ }\\!\\!\Delta\\!\\!\text{ }{{\text{T}}_{\text{b}}}\text{ = i m }{{\text{K}}_{\text{b}}}\text{ }$

We are given the following data:
Ebullioscopic constant of the solution is $\text{ }{{\text{K}}_{b}}\text{ }=\text{ }0.52\text{ kg Kmo}{{\text{l}}^{-1}}\text{ }$,
Weight of the compound $\text{ CuC}{{\text{l}}_{\text{2}}}\text{ }$ given,${{w}_{1}}\text{ = 13}\text{.44 gm }$
Weight of water (solvent), ${{w}_{2\text{ }}}=\text{ 1 kg = 1000 g }$
The molecular weight $\text{ CuC}{{\text{l}}_{\text{2}}}\text{ }$ is $134.4\text{ gm }$

We are interested to find the elevation in the boiling point $\text{ }\\!\\!\Delta\\!\\!\text{ }{{\text{T}}_{\text{b}}}\text{ }$.
We will solve this question by considering the van’t Hoff factor ‘i’.The elevation in the boiling point is the difference $\text{ }\\!\\!\Delta\\!\\!\text{ }{{\text{T}}_{\text{b}}}\text{ }$between the boiling point of the solution (solvent + solute ) and that of the pure solvent.
The $\text{ }\\!\\!\Delta\\!\\!\text{ }{{\text{T}}_{\text{b}}}\text{ }$ is expressed as follows,
$\text{ }\\!\\!\Delta\\!\\!\text{ }{{\text{T}}_{\text{b}}}\text{ = }{{\text{T}}_{\text{b}}}\text{ }-\text{ }{{\text{T}}_{\text{b}}}^{\text{0}}\text{ }$

The van’t Hoff factor can be introduced in calculating the observed value of the difference in the boiling point. The $\text{ }\\!\\!\Delta\\!\\!\text{ }{{\text{T}}_{\text{b}}}\text{ }$ is given as follows,
$\text{ }\\!\\!\Delta\\!\\!\text{ }{{\text{T}}_{\text{b}}}\text{ = i m }{{\text{K}}_{\text{b}}}\text{ }$
Where $\text{ }\\!\\!\Delta\\!\\!\text{ }{{\text{T}}_{\text{b}}}\text{ }$ is the difference in the boiling point
‘i’ is the Van’t Hoff factor
m is the concentration of the solute in terms of molality and ${{\text{K}}_{\text{b}}}\text{ }$ is the ebullioscopic constant.

The $\text{ CuC}{{\text{l}}_{\text{2}}}\text{ }$ dissociated into its ions in the solvent. The solute dissociates thus we are interested to find out the value of the number of ions in the solution. The $\text{ CuC}{{\text{l}}_{\text{2}}}\text{ }$ dissociates as follows,
$\text{ CuC}{{\text{l}}_{\text{2}}}\text{(s) }\to \text{ C}{{\text{u}}^{\text{2+}}}(l)\text{ + 2C}{{\text{l}}^{-}}(aq)\text{ }$
The $\text{ CuC}{{\text{l}}_{\text{2}}}\text{ }$ produces a total of 3 ions in the solution. Thus , $\text{ i = 3 }$.

Let's calculate the number of moles of$\text{ CuC}{{\text{l}}_{\text{2}}}\text{ }$,
We know that number of moles is the ratio of weight to the molar mass, thus
$\text{ no}\text{.of moles of CuC}{{\text{l}}_{\text{2}}}\text{ }=\text{ }\dfrac{\text{weight}}{\text{Molar mass}}\text{ = }\dfrac{13.44}{134.4}\text{ = 0}\text{.1 moles}$
Thus the number of moles $\text{ CuC}{{\text{l}}_{\text{2}}}\text{ }$ is equal to the$\text{0}\text{.1}$.

Therefore, the molality of the solution would be equal to,
$\text{ Molality (m) = }\dfrac{\text{no}\text{. of moles of CuC}{{\text{l}}_{\text{2}}}}{\text{Weight of solvent in Kg}}\text{ = }\dfrac{\text{0}\text{.1}}{\text{1}}\text{ = 0}\text{.1 m}$
Therefore, the molality of the solution is equal to $\text{0}\text{.1}$ molal.
Now, let’s calcite the elevation in the boiling point. Substitute the values in the equation. We have,

& \text{ }\\!\\!\Delta\\!\\!\text{ }{{\text{T}}_{\text{b}}}\text{ = i m }{{\text{K}}_{\text{b}}}\text{ } \\\ & \therefore \text{ }\\!\\!\Delta\\!\\!\text{ }{{\text{T}}_{\text{b}}}\text{ = (3) (0}\text{.1) (0}\text{.52) = 0}\text{.156 }\simeq \text{ 0}\text{.16 } \\\ \end{aligned}$$ Therefore, the elevation in the boiling point of the copper chloride solution is equal to the $\text{0}\text{.16}$. **So, the correct answer is “Option C”.** **Note:** The inorganic salt like $\text{ CuC}{{\text{l}}_{\text{2}}}\text{ }$ dissociate into the solution. In such a case, the number of effective particles increases, and therefore, the colligative properties like osmotic pressure, depression in freezing point, and elevation in boiling point are much higher than those of the calculated for the undissociated molecule. If we do not consider the dissociation of $\text{ CuC}{{\text{l}}_{\text{2}}}\text{ }$ then the $\text{ }\\!\\!\Delta\\!\\!\text{ }{{\text{T}}_{\text{b}}}\text{ }$ value would be equal to,$\text{ }0.052\text{ }$.$\text{ }\Delta {{\text{T}}_{\text{b}}}_{\text{(undissociated)}}=0.052\text{ }<\Delta {{\text{T}}_{\text{b}}}_{\text{(Dissociated)}}\text{ = 0}\text{.16 }$