Chapter 14, Problem 103A

Ice Cream A rock salt (NaCl), ice, and water mixture isused to cool milk and cream to make homemade icecream. How many grams of rock salt must be added towater to lower the freezing point by 10.0°C?

Interpretation Introduction

Interpretation:

The amount of rock saltis to be calculated.

Concept introduction:

For electrolytic solutes, the depression of the freezing point is directly proportional to molalityof the solution.

Freezing point depression $\Delta {\mathrm{T}}_f=i\times K_f\times m$

$\Delta {\mathrm{T}}_f$ is the Depression of the freezing point = the freezing point difference between a solution and the pure solvent.

$i$ is the Van’t Hoff factor and for strong electrolyte $i$ is equal to number of dissociate ions.

$K_f$ is the molalfreezing point depression constant.

$m$ is the molality of the solution.

Answer to Problem 103A

157 gm of NaCl must be added in 1kg of solvent.

Explanation of Solution

Given information:

$\Delta {\mathrm{T}}_f=10.0°\mathrm{C}$

Write the expression for depression of the freezing point.

$\Delta {\mathrm{T}}_f=i\times K_f\times m$

NaClis strong electrolyte, it dissociates completely in the solution to give two ions (Na+ and Cl- ion). So, i = 2

$\begin{array}{l}\Delta {\mathrm{T}}_f=i\times K_f\times m\\ MolalityofNaClsolution=\frac{\Delta {\mathrm{T}}_f}{i\times K_f}\\ =\frac{10.0°C}{2\times 1.86°C/m}=2.69m\end{array}$

Calculate the number of moles of $\mathrm{NaCl}$.

$\begin{array}{l}MolalityofNaClsolution=\frac{Numberofmolesof NaCl}{ Mass of water(inKg)}\\ Numberofmolesof NaCl=2.69m\times 1kg=2.69mole\end{array}$

$\begin{array}{l}MassofNaCl=NumberofmolesofNaCl\times MolecularweightofNaCl\\ =2.69mol\times 58.44gm/mol=157gm\end{array}$

(Note: Molecular weight of NaCl = 58.44 gm/mol)

Thus, 157 gm of NaCl must be added in 1kg of solvent.