Chapter 14, Problem 82A

Interpretation Introduction

Interpretation:

The molality of a solution have to be calculated.

Concept introduction:

Molality: Molality is defined as the number of moles of solute perkg of the solvent.

In expression,

$\mathrm{Molality}ofthesolution=\frac{Numberofmolesof the solute}{ Mass of the solvent(inKg)}$

$\mathrm{Number}ofmolesofsolute=\frac{Massofthesolute}{Molecularweightofthesolute}$

Answer to Problem 82A

The molality of a solution containing 30.0 g of naphthalene dissolved in 500.0 g of toluene is 0.468m.

Explanation of Solution

Given information:

Mass of naphthalene = 30.0 gm

Mass of the solvent toluene = 500.0 gm = 0.5 kg

Write the expression for Molality.

$\mathrm{Molality}ofthesolution=\frac{Numberofmolesof the solute}{ Mass of the solvent(inKg)}$

Calculate the number of moles of naphthalene.

$\begin{array}{l}Numberofmolesofnaphthalene=\frac{Massofnaphthalene}{Molecularweightofnaphthalene}\\ Numberofmolesofnaphthalene=\frac{30.0gm}{128gm/mol}=0.234mole\end{array}$

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

$\begin{array}{l}Molalityofnaphthalenesolutionintoluene=\frac{Numberofmolesofnaphthalene}{ Mass of toluene}\\ =\frac{0.234mole}{ 0.5kg}=0.468m\end{array}$

Thus, the molality of a solution containing 30.0 g of naphthalene dissolved in 500.0 g of toluene is 0.468m.