Chapter 13, Problem 23QP

Interpretation Introduction

Interpretation: The molarity and molality of the ${\text{NH}}_3$ solution is to be calculated.

Concept introduction:

The number of moles of solute dissolved in per $\text{kg}$ of a solvent is molality.

The expression to calculate molality $\left(m\right)$ is:

$m=\frac{\text{moles of solute}}{\text{mass of solvent}\left(\text{in kg}\right)}$` ……. (1)

The expression to calculate moles $\left(n\right)$ is:

$n=\frac{\text{mass}}{\text{molar mass}}$ …… (2)

Molarity $\left(M\right)$ is the number of moles of solute present in per liter of a solution.

The formula to calculate molarity is:

$M=\frac{\text{moles of solute}}{\text{liters of solution }}$ …… (3)

Conversion factor for $\text{g}$ to $\text{kg}$ is: $\frac{1\text{ g}}{{\text{10}}^{-3}\text{ kg}}$

Conversion factor for $\text{mL}$ to $\text{L}$ is: $\frac{{\text{10}}^{-3}\text{ L}}{\text{ 1 mL}}$