Chapter 12, Problem 16PE

Interpretation Introduction

Interpretation:

The molarity and molality of the solution are to be calculated if one gram of aluminum nitrate is dissolved in $\text{1 L}$ of water at $20°\text{C}$. The density of water at this temperature is $0.9982{\text{g/cm}}^3$ and the density of the resulting solution is $0.9989{\text{g/cm}}^3$.

Concept Introduction:

The number of moles of solute which are dissolved in one kilogram of the solvent is known as molality. It can be determined as:

$m=\frac{\text{mol of solute}\left(n\right)}{\text{kg of solvent}}$

The number of moles present in one liter of solution is known as molarity of the solution. It is calculated as:

$M=\frac{n}{V}$

Here, $M$ is the molarity, $V$ is the volume of solution in liters, and $n$ is the number of moles of solute.

The number of moles of solute is calculated by dividing the mass of solute by the molar mass of the solute as:

$n=\frac{m}{M_{\text{w}}}$

Here, $m$ is the mass of solute and $M_{\text{w}}$ is the molar mass of the solute.

The term percentage by mass $\left(\%w/w\right)$ shows the weight of solute dissolved in $100\text{ g}$ of solution. The percentage can be calculated as:

$\%=\frac{\text{ mass of solute }}{\text{ mass of solution}}\times 100$

Density is determined as:

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

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