Chapter 13, Problem 120AP

Interpretation Introduction

Interpretation:

The relation between molality and molarity is to be derived, and the fact that, for dilute aqueous solutions, molality is equal to molarity is to be proved.

Concept introduction:

Molality is defined as the ratio of the number of moles of the solute to the mass of the solvent (in kilograms). It is expressed as follows:

$m=\frac{n}{M}$ …… (3)

Here, $m$ is the molality, $n$ is the moles of the solute, and $M$ is the mass of the solvent (in kilograms).

Molarity is defined as the ratio of the number of moles of the solute to the volume of the solution (in liters). It is expressed as follows:

$m=\frac{n}{V}$ …… (3)

Here, $m$ is the molality, $n$ is the moles of the solute, and $V$ is the volume of the solution(in liters).

Density is defined as the ratio of mass to volume. It is expressed as follows:

$d=\frac{m}{V}$

Here, $d$ is the density, $m$ is the mass, and $V$ is the volume.

Answer to Problem 120AP

Solution:

(a)

The relation between molality and molarity has been derived.

(b)

For dilute solutions, molality and molarity are equal.

Explanation of Solution

a)Drive the equation relating the molality and molarity of a solution

The mass of the solvent (in kilograms) is calculated as follows:

$Massofsolvent(kg)=\left[massof\text{solution }\left(g\right)-massofsolute\left(g\right)\right]\times \frac{1kg}{1000g}$

Or

$Massofsolvent(kg)=\frac{massof\text{solution }\left(g\right)-massofsolute\left(g\right)}{1000}$…… (1)

Consider $1\text{ L}$ or $1000\text{ mL}$ of solution.

Density is calculated as follows:

$d=\frac{m}{V}$

Rearrange the above equation for the calculation of mass as follows:

$m=\text{ d }\times \text{V}$

$massof\text{solution}=\text{d}\left(\frac{g}{mL}\right)\times 1000mL$

Calculate the mass of the solution from the molarity and its molar mass, as follows:

$M=\frac{\text{moles of solute}}{\text{volume of solution in liters}}$…… (2)

Number of moles is calculatedas follows:

$\text{Number of moles}=\frac{\text{mass }}{\text{molar mass}}\text{ or }\frac{m}{M}$…… (3)

By substituting equation (3) in equation (2), we will get:

$M=\frac{\text{m/}M}{\text{volume of solution in liters}}$

Rearrange the above equation for the calculation of mass as follows:

$massofsolute=\text{M}\left(\frac{mol}{L}\right)\times 1L\times M\left(\frac{g}{mol}\right)$

Substituting these expressions into equation (1),

$Massofsolvent(kg)=\frac{(\text{d})(1000)-\text{M}M}{1000}$

Or

$Massofsolvent(kg)=\text{d}-\frac{\text{M}M}{1000}$…… (4)

Molality is defined as the number of moles of the solute divided by the mass of the solvent (in kilograms).

It is expressed as follows:

$m=\frac{\text{moles of solute}}{\text{mass of solvent in kg}}$

Rearrange the above equation for the calculation of mass as follows:

$Massofsolvent(kg)=\frac{n}{m}$…… (5)

Consider $1\text{ L}$ of solution. So, the moles of the solute (n) become equal to itsmolarity (M). Then, equation (5)becomes:

$\text{Mass of}solvent(kg)=\frac{\text{M}}{\text{m}}$

Substituting the above equation back into equation (4) gives the following equation:

$\frac{\text{M}}{\text{m}}=\text{d}-\frac{\text{M}M}{1000}$

Taking the inverse of both sides of the equation gives the following equation:

$\frac{\text{m}}{\text{M}}=\frac{1}{\text{d}-\frac{\text{M}M}{1000}}$

or

$\text{m}=\frac{\text{M}}{\text{d}-\frac{\text{M}M}{1000}}$

Hence, the above equation is the relation between the molality of a solution to its molarity.

b) For any aqueous solution, molality is equal to molarity.

The density of water is approximately $1\text{ g/mL}$. Therefore, in dilute solutions, the density of a dilute aqueous solution is approximately $1\text{ g/mL}$.

In dilute solutions, $\text{d}>>\frac{\text{M}M}{1000}$.

Consider a $0.010\text{ M}$ NaCl solution.

$\begin{array}{c}\frac{\text{M}M}{1000}=\frac{(0.010\text{mol/L})(58.44\text{g/mol})}{1000}\\ =5.8\times {10}^{-4}\text{g/L}<<1\end{array}$

The derived equation reduces to the equation given below:

$\text{m}\approx \frac{\text{M}}{\text{d}}$

When the density becomes equal to $1\text{ g/mL}$, that is, $d\approx 1\text{ g/mL}$, the molality becomes equal to molarity $\left(m\approx M\right)$.