Chapter 14, Problem 101A

Cooking A cook prepares a solution for boiling by adding12.5 g of NaCl to a pot holding 0.750 L of water. Atwhat temperature should the solution in the pot boil?Use Table 14.5 for needed data.

Interpretation Introduction

Interpretation:

The boiling point of a solution is to be calculated.

Concept introduction:

For electrolyte solutes, the elevation of the boiling point is directly proportional to molalityof the solution.

Boiling point elevation $\Delta {\mathrm{T}}_b=i\times K_b\times m$

Where,

$\Delta {\mathrm{T}}_b$ is the elevation of boiling point = the boiling point difference between a solution and the pure solvent.

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

$K_b$ is the molal boiling point elevation constant.

$m$ is the molality of the solution.

Answer to Problem 101A

The boiling point of the solution is 100.29 °C.

Explanation of Solution

Given information:

Mass of NaCl = 12.5 g Volume of water = 750 mL

Write the expression for the number of moles of NaCl.

$\begin{array}{c}NumberofmolesofNaCl=\frac{MassofNaCl}{MolecularweightofNaCl}\\ =\frac{12.5gm}{58.44gm/mol}\\ =0.214mole\end{array}$

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

Calculate the mass of water.

$\begin{array}{c}\mathrm{Mass}ofwater=\text{ Volume}ofwater\times Densityofwater\\ =750\mathrm{mL}\times 1\mathrm{gm}/mL\\ =750\mathrm{gm}\\ =0.75\mathrm{kg}\end{array}$

Write the expression for the Molality.

$\begin{array}{l}MolalityofNaClsolution=\frac{Numberofmolesof NaCl}{ Mass of water(inKg)}\\ =\frac{0.214mole}{ 0.75kg}=0.285m\end{array}$

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

$\begin{array}{l}\Delta T_b=i\times K_b\times m\\ =2\times 0.512°C/m\times 0.285m\\ =0.292°C\end{array}$

Thus, the boiling point of the solution is:

$\begin{array}{c}{T}_b=100°\mathrm{C}+0.292°\mathrm{C}\\ =100.29°\mathrm{C}\end{array}$

Hence the solution is $T_b=100.29°\mathrm{C}$.