Chapter 13, Problem 13.79SP

Interpretation Introduction

(a)

Interpretation:

The mole fraction of lactose in 0.335 M solution needs to be determined, if the density of aqueous solution is 1.0432 g/mL at 20°C.

Concept introduction:

Solution stoichiometry involves the calculation of concentration of solutions in the given conditions of volumes, moles etc. There are various ways to calculate the concentration of solutions such as molarity, molality, mole fraction, ppm, ppb etc. Mole fraction is the ratio of moles of substance and total moles in the solution or mixture. One ppm stands for part per million or milligrams per liter (mg/L) whereas parts per billion (ppb) is one part in 1 billion.

Molarity represents the moles of solute dissolve in per liter of solution. The mathematical expression of molarity is:

$\text{Molarity=}\frac{\text{Moles of solute}}{\text{Volume of solution (L)}}$

Mole fraction of any component in a solution can be calculated as:

$\text{Mole fracation of A =}\frac{\text{Moles of A}}{\text{moles of A + moles of B}}$

Interpretation Introduction

(b)

Interpretation:

Themass percentage of lactose in 0.335 M solution needs to be determined, if the density of aqueous solution is 1.0432 g/mL at 20°C.

Concept introduction:

Solution stoichiometry involves the calculation of concentration of solutions in the given conditions of volumes, moles etc. There are various ways to calculate the concentration of solutions such as molarity, molality, mole fraction, ppm, ppb etc. Mole fraction is the ratio of moles of substance and total moles in the solution or mixture. One ppm stands for part per million or milligrams per liter (mg/L) whereas parts per billion (ppb) is one part in 1 billion.

Molarity represents the moles of solute dissolve in per liter of solution. The mathematical expression of molarity is:

$\text{Molarity=}\frac{\text{Moles of solute}}{\text{Volume of solution (L)}}$

Mole fraction of any component in a solution can be calculated as:

$\text{Mole fracation of A =}\frac{\text{Moles of A}}{\text{moles of A + moles of B}}$

Interpretation Introduction

(c)

Interpretation:

Themolality of lactose in 0.335 M solution needs to be determined, if the density of aqueous solution is 1.0432 g/mL at 20°C.

Concept introduction:

Solution stoichiometry involves the calculation of concentration of solutions in the given conditions of volumes, moles etc. There are various ways to calculate the concentration of solutions such as molarity, molality, mole fraction, ppm, ppb etc. Mole fraction is the ratio of moles of substance and total moles in the solution or mixture. One ppm stands for part per million or milligrams per liter (mg/L) whereas parts per billion (ppb) is one part in 1 billion.

Molality represents the moles of solute dissolve in per kg of solvent. The mathematical expression of molality is:

$\text{Molality =}\frac{\text{moles of solute}}{\text{mass of solvent (kg)}}$

Mole fraction of any component in a solution can be calculated as:

$\text{Mole fracation of A =}\frac{\text{moles of A}}{\text{moles of A + moles of B}}$