Question

$15.0\,{\text{g}}$ of an unknown molecular material was dissolved in $450\,{\text{g}}$ of water. The resulting solution was found to freeze at $- 0.34{\,^{\text{o}}}{\text{C}}$. What is the molar mass of this material (${{\text{k}}_{\text{f}}}$ for water is $= 1.86\,{\text{K kg mo}}{{\text{l}}^{ - 1}}$ ).

Answer 1

The depression in freezing point is the product of the freezing point depression constant and molality. The formula used in depression in freezing point is given as: $\Delta {T_f}\, = {K_f}.\,m$

Complete step by step answer:
The formula of freezing point depression is as follows:
$\Delta {T_f}\, = {K_f}.\,m$
Where,
$\Delta {T_f}$ is the depression in freezing point.
${K_f}$ is the freezing point depression constant.
Determine the depression in freezing point as follows:
The freezing point of pure water is ${0^{\text{o}}}{\text{C}}$.
$\Delta {T_f}\, = \,{T_{{\text{water}}}} - {T_{solution}}$
Substitute $0^\circ$for freezing point of pure water and $- 0.34{\,^{\text{o}}}{\text{C}}$ for freezing point of solution.
$\Delta {T_f}\, = \,0{\,^{\text{o}}}{\text{C}}\, - \left( { - 0.34{\,^{\text{o}}}{\text{C}}} \right)$
$\Delta {T_f}\, = \,{0.34^{\text{o}}}{\text{C}}$
Rearrange the formula for molality as follows:
$m\, = \,\dfrac{{\Delta {T_f}}}{{{K_f}}}$
Substitute $1.86\,{\text{K kg mo}}{{\text{l}}^{ - 1}}$ for freezing point depression constant and ${0.34^{\text{o}}}{\text{C}}$ for depression in freezing point.
$m\, = \,\dfrac{{0.34\,{\text{K}}}}{{1.86\,{\text{K kg mo}}{{\text{l}}^{ - 1}}}}$
$m\, = \,0.183\,{\text{mol}}\,\,{\text{k}}{{\text{g}}^{ - 1}}$
So, the molality of the unknown material solution is $0.183\,{\text{mol}}\,\,{\text{k}}{{\text{g}}^{ - 1}}$.
Determine the amount of unknown material dissolved in $450\,{\text{g}}$ of water as follows:
Convert the amount of water from gram to kilogram as follows:
$1000{\text{g = }}\,{\text{1}}\,{\text{kg}}$
$450{\text{g = }}\,0.450\,{\text{kg}}$
Molality is defined as the amount of solute (unknown material) dissolved in an amount of solvent (water). So, $0.183\,{\text{mol}}\,$ unknown material is dissolved in ${\text{1}}\,{\text{kg}}$ of water.
The formula of molarity is as follows:
${\text{Molality}}\,{\text{ = }}\dfrac{{{\text{Mole}}\,{\text{of}}\,{\text{solute}}}}{{{\text{kg}}\,{\text{of}}\,{\text{solvent}}}}$
Substitute $0.450\,{\text{kg}}$ for mass of solvent and $0.183\,{\text{mol}}\,\,{\text{k}}{{\text{g}}^{ - 1}}$ for molality.
${\text{0}}{\text{.183}}\,{\text{mol}}\,\,{\text{k}}{{\text{g}}^{ - 1}}\,{\text{ = }}\dfrac{{{\text{Mole}}\,{\text{of}}\,{\text{solute}}}}{{{\text{0}}{\text{.450}}\,{\text{kg}}}}$
${\text{mole}}\,{\text{of}}\,{\text{solute}} = 0.183\,{\text{mol}}\,\,{\text{k}}{{\text{g}}^{ - 1}} \times 0.450\,{\text{kg}}\,$
${\text{Mole}}\,{\text{of}}\,{\text{solute}} = 0.0823\,{\text{mol}}\,\,$
So, the mole of unknown material is $0.0823\,{\text{mol}}\,$.
Use the mole formula to determine the molar mass of unknown material.
${\text{Mole}}\,{\text{ = }}\,\dfrac{{{\text{Mass}}}}{{{\text{Molar}}\,{\text{mass}}}}$
The mass of unknown material is $15.0\,{\text{g}}\,$.
Substitute $0.0823\,{\text{mol}}\,$for mole and $15.0\,{\text{g}}\,$ for mass.
$0.0823\,\,{\text{mol}}\,\,{\text{ = }}\,\dfrac{{15.0\,{\text{g}}}}{{{\text{Molar }}\,{\text{mass}}\,}}$
${\text{Molar }}\,{\text{mass}}\,{\text{ = }}\,\dfrac{{15.0\,{\text{g}}}}{{0.0823\,{\text{mol}}\,}}$
${\text{Molar mass}} = \,{\text{182}}{\text{.26 g /mol}}$
Therefore, the molar mass of the material is ${\text{182}}{\text{.3}}\,{\text{g/mol}}$.

Note: When a solute is added to the pure solvent, the freezing point of the solution decreases which is known as the depression in the freezing point. The freezing point of the pure solvent at a temperature is defined as the freezing point constant.