Chapter 16, Problem 16.11P

(a)

Interpretation Introduction

Interpretation:

The colligative molality of $1.0$ mole $MgS{O}_4$ dissolved in $1000$ grams of water has to be calculated.

Concept Introduction:

Molality:

Molality is defined as the number of moles of solute per $1000g$ or one kilogram of solvent.  Mathematically, it can be represented as given below.

  $Molality=\frac{Molesofsolute}{1kgofsolvent}$

The colligative molality is the product of van’t Hoff factor and molality.  Mathematically, it can be represented as given below.

  $m_c=i\times m$

Van’t Hoff factor (i) is the total number of ions produced in the solution.

(b)

Interpretation Introduction

Interpretation:

The colligative molality of $1.0$ mole $Cu{\left(N{O}_3\right)}_2$ dissolved in $1000$ grams of water has to be calculated.

Concept Introduction:

Refer to part (a).

(c)

Interpretation Introduction

Interpretation:

The colligative molality of $1.0$ mole $C{H}_{3}C{H}_{2}OH$ dissolved in $1000$ grams of water has to be calculated.

Concept Introduction:

Refer to part (a).

(d)

Interpretation Introduction

Interpretation:

The colligative molality of $1.0$ mole $A{l}_2{\left(S{O}_4\right)}_3$ dissolved in $1000$ grams of water has to be calculated.

Concept Introduction:

Refer to part (a).