Question

Assume that each of the ions in calcium chloride, \({\bf{CaC}}{{\bf{l}}_{\bf{2}}}\), has the same effect on the freezing point of water as a nonelectrolyte molecule. Calculate the freezing point of a solution of 0.724 g of \({\bf{CaC}}{{\bf{l}}_{\bf{2}}}\)in 175 g of water.

Short answer

The freezing point of a solution is−0.208 °C.

Step-by-step solution

Definition of Freezing point

The freezing point is the temperature at which a liquid changes into a solid, at normal atmospheric pressure.

\({\bf{\Delta }}{{\bf{T}}_{\bf{f}}}{\bf{ = }}{{\bf{K}}_{\bf{f}}}{\bf{m}}\)

Explanation

Mass of\({\rm{CaC}}{{\rm{l}}_{\rm{2}}}\)= 0.724 g

Molar Mass of\({\rm{CaC}}{{\rm{l}}_{\rm{2}}}\)= 111 g/mole

\({\rm{Number of Moles = }}\frac{{{\rm{0}}{\rm{.724}}}}{{{\rm{111}}}}{\rm{ = 0}}{\rm{.0065mole}}\)

The number of moles of ions present in the solution using the number of moles of ions in 1 mole of \({\rm{CaC}}{{\rm{l}}_{\rm{2}}}\)as the conversion factor (3 mole ions/1 mole \({\rm{CaC}}{{\rm{l}}_{\rm{2}}}\)).

\({\rm{Number of Moles = 0}}{\rm{.0065 \times 3 = 0}}{\rm{.019mole}}\)

Mass of Solvent = 175 g = 0.175 kg

\({\rm{Molarity = }}\frac{{{\rm{Number of Moles}}}}{{{\rm{Volume of Solvent }}\left( {{\rm{Kg}}} \right)}}\)

\({\rm{Molality = }}\frac{{{\rm{0}}{\rm{.019mole}}}}{{{\rm{0}}{\rm{.175kg}}}}{\rm{ = 0}}{\rm{.108m}}\)

Freezing point depression constant = 1.86 °C/m

\(\begin{aligned}{{}{}}{{\rm{\Delta }}{{\rm{T}}_{\rm{f}}}{\rm{ = }}{{\rm{K}}_{\rm{f}}}{\rm{m}}}\\{{\rm{\Delta }}{{\rm{T}}_{\rm{f}}}{\rm{ = 1}}{\rm{.86 \times 0}}{\rm{.108 = 0}}{\rm{.208}}}\\{{\rm{Freezing point of solution = }}\left( {{\rm{0 - 0}}{\rm{.208}}} \right){\rm{ ^\circ C = - 0}}{\rm{.208 ^\circ C}}}\end{aligned}\)