Question

${\text{CaC}}{{\text{l}}_{\text{2}}}$ is preferred over ${\text{NaCl}}$ for clearing ice on roads on particularly in very cold countries. This is because:
A) ${\text{CaC}}{{\text{l}}_{\text{2}}}$ is not soluble in ${{\text{H}}_{\text{2}}}{\text{O}}$ than ${\text{NaCl}}$.
B) ${\text{CaC}}{{\text{l}}_{\text{2}}}$ is hygroscopic but ${\text{NaCl}}$ is not.
C) Eutectic mixture of ${\text{CaC}}{{\text{l}}_{\text{2}}}{\text{/}}{{\text{H}}_{\text{2}}}{\text{O}}$ freezes at $ - {55^ \circ }{\text{C}}$ while that of ${\text{NaCl/}}{{\text{H}}_{\text{2}}}{\text{O}}$ freezes at $ - {18^ \circ }{\text{C}}$.
D) ${\text{NaCl}}$ makes the road slippery but ${\text{CaC}}{{\text{l}}_{\text{2}}}$ does not.

Answer 1

Hint: ${\text{CaC}}{{\text{l}}_{\text{2}}}$ dissociates into three ions, one chloride ion and two calcium ions. Thus, the van’t Hoff factor for ${\text{CaC}}{{\text{l}}_{\text{2}}}$ is $3$. ${\text{NaCl}}$ dissociates into two ions, one chloride ion and one sodium ion. Thus, the van’t Hoff factor for ${\text{NaCl}}$ is $2$.

Formula used: $\Delta {T_f} = {K_f} \times m \times i$

Complete step by step answer:
A mixture of substances that melts at a temperature lower than the melting points of its constituent substances is known as eutectic mixture.
The freezing point of pure solvent decreases when a non-volatile solute is added to it. Thus, when ${\text{CaC}}{{\text{l}}_{\text{2}}}$ or ${\text{NaCl}}$ is added to water, the freezing point of water decreases.
The formula for the depression in freezing point is,
$\Delta {T_f} = {K_f} \times m \times i$
Where $\Delta {T_f}$ is the depression in freezing point,
${K_f}$ is the freezing point depression constant,
m is the molality of the solution,
i is the van’t Hoff factor.
From the formula for the depression in freezing point, we have,
$\Delta {T_f} \propto i$
Thus, the depression in freezing point is directly proportional to the van’t Hoff factor.
${\text{CaC}}{{\text{l}}_{\text{2}}}$ dissociates into three ions, one chloride ion and two calcium ions. Thus, the van’t Hoff factor for ${\text{CaC}}{{\text{l}}_{\text{2}}}$ is $3$. ${\text{NaCl}}$ dissociates into two ions, one chloride ion and one sodium ion. Thus, the van’t Hoff factor for ${\text{NaCl}}$ is $2$.
The number of individual ions any ionic compound dissociates into is known as the van’t Hoff factor. The van’t Hoff factor for ${\text{CaC}}{{\text{l}}_{\text{2}}}$ is greater than that of ${\text{NaCl}}$. Thus, the freezing point of water when ${\text{CaC}}{{\text{l}}_{\text{2}}}$ is added to it is much lower than when ${\text{NaCl}}$ is added to it.
Thus, the eutectic mixture of ${\text{CaC}}{{\text{l}}_{\text{2}}}{\text{/}}{{\text{H}}_{\text{2}}}{\text{O}}$ freezes at $ - {55^ \circ }{\text{C}}$ while that of ${\text{NaCl/}}{{\text{H}}_{\text{2}}}{\text{O}}$ freezes at $ - {18^ \circ }{\text{C}}$.
Thus, the correct option is (C) eutectic mixture of ${\text{CaC}}{{\text{l}}_{\text{2}}}{\text{/}}{{\text{H}}_{\text{2}}}{\text{O}}$ freezes at $ - {55^ \circ }{\text{C}}$ while that of ${\text{NaCl/}}{{\text{H}}_{\text{2}}}{\text{O}}$ freezes at $ - {18^ \circ }{\text{C}}$.

Hence the correct option is (C).

Note: The ability of any substance to dissolve is known as solubility. The solubility of ${\text{CaC}}{{\text{l}}_{\text{2}}}$ in water is more than the solubility of ${\text{NaCl}}$.
Thus, option (A) is not correct.
A substance that can absorb moisture from the air is known as hygroscopic. Both ${\text{CaC}}{{\text{l}}_{\text{2}}}$ and ${\text{NaCl}}$ can absorb moisture from air and are hygroscopic in nature.
Thus, option (B) is not correct.
Both the ${\text{CaC}}{{\text{l}}_{\text{2}}}$ and ${\text{NaCl}}$ solutions can make the roads slippery.
Thus, option (D) is not correct.