Chapter 16, Problem 16.70P

Interpretation Introduction

Interpretation:

The freezing points of wine and vodka have to be calculated.

Concept Introduction:

Freezing point depression:

The lowering of the vapor pressure of a solvent by a solute leads to a lowering of the freezing point of the solution relative to that of the pure solvent.  This effect is called as the freezing point depression.  The mathematical relationship is given below.

  $\Delta T_f=T_f{}^{\circ }-T_f=K_f{m}_c$

Where, $\Delta T_f$ is freezing point depression, $T_f$ is the freezing point of solution, $T_f{}^{\circ }$ is the freezing point of pure solvent, $K_f$ is the freezing point depression constant and $m_c$ is the colligative molality.

Answer to Problem 16.70P

The freezing points of wine and vodka are $-4.4^{o}Cand-21^{o}C$ respectively.

Explanation of Solution

The value of $K_f$ for water is $1.86K.m_c{}^{-1}$ and its freezing point is $0^{\circ }C.$

Wine and vodka are about $12\%and40\%$ of ethanol by volume.  That means, wine contains $12mL$ of ethanol and $88mL$ of water.  Similarly, vodka contains $40mL$ of ethanol and $60mL$ of water.

Given that, the densities of ethanol and water are $0.79g.m{L}^{-1}and1.00g.m{L}^{-1}$ respectively.

The mass of ethanol and water in wine and vodka can be calculated as given below.

  $\begin{array}{l}Mass=Density\times Volume\\ \\ Massofethanolinwine=0.79g.m{L}^{-1}\times 12mL=9.48g\\ \\ Massofwaterinwine=1.00g.m{L}^{-1}\times 88mL=88g\\ \\ Massofethanolinvodka=0.79g.m{L}^{-1}\times 40mL=31.6g\\ \\ Massofwaterinvodka=1.00g.m{L}^{-1}\times 60mL=60g\end{array}$

The colligative molality of wine and vodka can be calculated as given below.

  $\begin{array}{l}Molality=\frac{Moles}{kgofsolvent}\\ \\ Molalityofwine=\frac{\frac{Massofethanol}{Molarmassofethanol}}{kgofsolvent}\\ \\ Molalityofwine=\frac{\frac{9.48g}{46.07g.m{L}^{-1}}}{88g\times \left(\frac{1kg}{1000g}\right)}\\ \\ Molalityofwine=2.34m\\ \\ Molalityofvodka=\frac{\frac{Massofethanol}{Molarmassofethanol}}{kgofsolvent}\\ \\ Molalityofvodka=\frac{\frac{31.6g}{46.07g.m{L}^{-1}}}{60g\times \left(\frac{1kg}{1000g}\right)}\\ \\ Molalityofvodka=11.43m\end{array}$

The colligative molality and molality are same for ethanol.

Now, using the freezing point depression expression, the freezing points of wine and vodka can be calculated.

For wine:

  $\begin{array}{l}\Delta T_f=T_f{}^{\circ }-T_f=K_f{m}_c=\left(1.86K.m_c{}^{-1}\right)\times \left(2.34m_c\right)\\ \\ 0^{\circ }C-\left(T_f\right)=4.4K\\ \\ T_f=-4.4^{o}C\end{array}$

For vodka:

  $\begin{array}{l}\Delta T_f=T_f{}^{\circ }-T_f=K_f{m}_c=\left(1.86K.m_c{}^{-1}\right)\times \left(11.43m_c\right)\\ \\ 0^{\circ }C-\left(T_f\right)=21K\\ \\ T_f=-21^{o}C\end{array}$

The value of $\Delta T_f$ in kelvins is equal to the value of $\Delta t_f$ in degrees Celsius.

Therefore, the freezing points of wine and vodka are $-4.4^{o}Cand-21^{o}C$ respectively.