Chapter 13, Problem 13.67SP

Interpretation Introduction

(a)

Interpretation:

The molality of a solution made by dissolving 0.655 mol of citric acid in 1.00 kg of water should be calculated.

Concept introduction:

The molarity of the solution is the ratio of the moles of solute and the weight of the solvent in kg. It is given by the formula,

$\text{Molality}=\frac{\text{Moles}\text{of}\text{solute}}{\text{Weight}\text{of}\text{solvent}\left(\text{kg}\right)}$

Interpretation Introduction

(b)

Interpretation:

The molality of the solution made by dissolving 0.135 mg of KBr in 5.00 mL of water should be calculated.

Concept introduction:

The molarity of the solution is the ratio of the moles of solute and the weight of the solvent in kg. It is given by the formula,

$\text{Molality}=\frac{\text{Moles}\text{of}\text{solute}}{\text{Weight}\text{of}\text{solvent}\left(\text{kg}\right)}$

Interpretation Introduction

(c)

Interpretation:

The molality of the solution made by dissolving 5.50 g of aspirin in 145 g of dichloromethane should be calculated.

Concept introduction:

The molarity of the solution is the ratio of the moles of solute and the weight of the solvent in kg. It is given by the formula,

$\text{Molality}=\frac{\text{Moles}\text{of}\text{solute}}{\text{Weight}\text{of}\text{solvent}\left(\text{kg}\right)}$