Chapter 14, Problem 102A

Interpretation Introduction

Interpretation:

The amount of naphthalene used to change the boiling point of ethanol have to be calculated.

Concept introduction:

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

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

Where,

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

$K_b$ is the molal boiling point elevation constant.

$m$ is the molality of the solution.

Answer to Problem 102A

703 gm of naphthalene used to change the boiling point of ethanol.

Explanation of Solution

Given information:

Mass of the solvent ethanol = 1 kg

Calculate the boiling point elevation.

$\begin{array}{c}\Delta {\mathrm{T}}_b=(85.2°C–78.5°C)\\ =6.7°C\end{array}$

Write the expression for boiling point elevation.

$\Delta {\mathrm{T}}_b=K_b\times m$

Calculate the molality of solution from above equation.

$\begin{array}{l}\Delta {\mathrm{T}}_b=K_b\times m\\ m=\frac{\Delta T_b}{{\mathrm{K}}_b}=\frac{6.7°C}{1.22°C/m}=5.49m\end{array}$

(Note: K_b of ethanol = 1.22 °C/m)

Calculate the number of moles of the naphthalene.

$\begin{array}{l}Molalityofnaphthalenesolution=\frac{Numberofmolesof naphthalene}{ Mass of ethanol(inKg)}\\ Numberofmolesof naphthalene=Molalityofnaphthalenesolution\times Mass of ethanol\\ =5.49m\times 1kg=5.49mole\end{array}$

Calculate the mass of naphthalene.

$\begin{array}{l}Numberofmolesofnaphthalene=\frac{Massofnaphthalene}{Molecularweightofnaphthalene}\\ Massofnaphthalene=Numberofmolesofnaphthalene\times Molecularweightofnaphthalene\\ =5.49mole\times 128gm/mol=703gm\end{array}$

(Note: Molecular weight of naphthalene = 128 gm/mol)

Thus, the amount of naphthalene is 703 gm.