Question

What mass of $CuC{l_2}$ to be dissolved in $100$ g of water to raise the boiling point by $0.16$ degree?
(Consider complete ionization, the molar mass of $CuC{l_2} = 134.5g{\left( {mol} \right)^{ - 1}}$ and ${K_b}$ of water $= 0.52K{m^{ - 1}}$
A.$1.379g$
B.$2.69g$
C.$4.035g$
D.$0.4035g$

Answer 1

The change in boiling point is caused when an addition of a salt takes place, it will increase the boiling point and the phenomenon can be called an elevation of boiling point. The ionization of salt has also been considered, as the value of van’t Hoff factor can also be included in calculating the elevation of boiling point.
Formula used:
$\Delta {T_b} = \dfrac{{i \times 1000 \times {K_b} \times {w_B}}}{{{M_b} \times {w_A}}}$
$\Delta {T_b}$ is change in boiling point
$i$ is van’t Hoff factor
${K_b}$ is molal elevation of constant $= 0.52K{m^{ - 1}}$
${w_A}$ is amount of solvent is $100$ gm
${M_B}$ is molar mass of solute $CuC{l_2} = 134.5g{\left( {mol} \right)^{ - 1}}$
${w_B}$ is mass of solute has to be determined
Thus, ${w_B} = \dfrac{{\Delta {T_b} \times {M_b} \times {w_A}}}{{i \times 1000 \times {K_b}}}$

Complete answer:
Given that copper chloride $CuC{l_2}$ is dissolved in $100$ g of water to raise the boiling point by $0.16$ degree.
Here, $0.16$ degree is $0.16$ kelvin
The van’t Hoff factor of copper chloride will be $3$.
Substitute all the values in the above formula
${w_B} = \dfrac{{0.16 \times 134.5 \times 100}}{{1000 \times 3 \times 0.52}}$
By simplifying, we will get
${w_B} = 1.379g$
Thus, the mass of copper chloride is $1.379g$.
In the given options, our answer matches with option A. Thus, option A is the correct one.

Note:
The van’t Hoff factor of copper chloride is $3$, as the complete dissociation of a salt means it was completely dissociated into ions. Copper chloride is an inorganic salt, upon dissociation it gives two ions of chloride and one ion of copper. Thus, the total ions formed are , which is van’t Hoff factor value.