Question

The solution to Problem 49 shows that to make homemade ice cream, temperatures ranging downward from $\mathbf{-}{\mathbf{3}}^{\mathbf{0}}\mathbf{\text{C}}$are needed. Ice cubes from a freezer have a temperature of about $\mathbf{-}{\mathbf{12}}^{\mathbf{0}}\mathbf{\text{C}}\left(+{10}^0\text{F}\right)\mathbf{,}$which is cold enough, but contact with the warmer ice cream mixture causes them to melt into liquid at ${\mathbf{0}}^{\mathbf{0}}\mathit{C}$which is too warm. To obtain a liquid that is cold enough, salt (NaCl)is dissolved in water, and ice is added to the saltwater. The salt lowers the freezing point of the water enough so that it can freeze the liquid inside the ice cream maker. The instructions for an ice cream maker say to add one-part salt to eight parts water (by mass). What is the freezing point of this solution (in degrees Celsius and degrees Fahrenheit)? Assume that the (NaCl) dissociates fully into ions and that the solution is ideal.

Short answer

The freezing-point of the solution is $-7.{96}^0\text{C} \text{or} 17.{67}^0\text{F}$.

Step-by-step solution

Given data

The temperature of the ice cube is $-{12}^0\text{C}\left(+{10}^0\text{F}\right)$.

The concept of depression of freezing point.

When the freezing point of a solution is lowered due to addition of a solute, it is known as depression of freezing point. The lowering of the freezing point depends only on the number of particles of the solute dissolved in the solution.

Calculation of the amount of water and NaCl

Number of moles present in sodium chloride can be determined as shown below.

$1000\text{g}\text{water}\times \frac{1\text{g salt}}{8\text{g water}}=125\text{g salt}$

$$\begin{gathered} \text{Number}\text{of}\text{moles}\text{of}\text{salt(NaCl) =}\frac{\text{mass in g}}{\text{molar mass in g/mol}} \\ \text{n(NaCl) =}\frac{125g}{58.44g/mol} \\ \text{=2.14mol} \end{gathered}$$

Calculation of molality

Molality is determined by the formula:

$$\begin{gathered} \text{molality(m) =}\frac{\text{number of moles of solute}}{\text{Mass of solvent in kg}} \\ \text{molality}\left(\text{m}\right)=\frac{2.14\text{mol}}{1\text{kg}} \\ =2.14\text{mol}/\text{kg} \end{gathered}$$

It is known that NaCl dissociates completely by forming${\text{Na}}^{\text{+}} \text{and} {\text{Cl}}^{\text{-}}$ ions.

Hence, molality of the solution is :

$\text{(m)}=2\times 2.14\text{mol}/\text{kg}$

.$\text{(m)}=4.28\text{mol}/\text{kg}$

Calculation of ΔTf

It can be calculated with the help of formula:

$\Delta T_f=K_{f}m$ ……(i).

Substituting the value of $K_f=1.86{\text{Kkgmol}}^{-1}$and $m=4.28\text{mol}/\text{kg}$in above equation.

$$\begin{gathered} \Delta T_f=\left(1.86{\text{Kkgmol}}^{-1}\right)(4.28\text{mol}/\text{kg}) \\ =7.96\text{K} \end{gathered}$$

Calculation of ΔTsolution 

The determination of$\Delta T_{\text{solution}}$ can be shown as follows:

$$\begin{gathered} \Delta T_{\text{solution}}=T_{\text{solvent}}-T_{\text{solution}} \\ \Delta T_{\text{solution}}=0.{00}^0\text{C}-7.{96}^0 \\ =-7.{96}^0\text{C} \end{gathered}$$

Calculation of freezing point in Farenheit

Freezing point can be determined by use of the formula:

$t^0\mathbf{F}=\frac{9^0\mathbf{F}}{5^0\text{C}}\left(t^0\text{C}\right)+{32}^0\mathbf{F}$

Substitute the value of given data in above equation.

$$\begin{gathered} t^0\mathrm{F}=\frac{9^0\mathrm{F}}{5^0\text{C}}\left(t^0\text{C}\right)+{32}^0\mathrm{F} \\ {t}^0\mathrm{F}=\frac{9^0\mathrm{F}}{5^0\text{C}}\left(-7.{96}^0\text{C}\right)+{32}^0\mathrm{F} \\ =17.{672}^0\mathrm{F} \\ \approx 17.{67}^0\mathrm{F} \end{gathered}$$