Chapter 3, Problem 3.68P

(a)

Interpretation Introduction

Interpretation:Mass in grams of calcium acetate should be calculated.

Concept introduction:The formula to evaluate moles from molarity is given as follows:

  $n=MV$

Where,

• $V$represents volume.
• $M$represents molarity.
• $n$represents number of moles.

The formula to convert mass in gram to moles is as follows:

  $\text{Mass}=\left(\text{Number of moles}\right)\left(\text{Molar mass}\right)$

(b)

Interpretation Introduction

Interpretation: Molarity of $500\text{ mL}$ of $21.1\text{ g KI}$ solution is to be calculated.

Concept introduction:The formula to evaluate moles from molarity is given as follows:

  $n=MV$

Where,

• $V$represents volume.
• $M$represents molarity.
• $n$represents number of moles.

The formula to convert mass in gram to moles is as follows:

  $\text{Mass}=\left(\text{Number of moles}\right)\left(\text{Molar mass}\right)$

(c)

Interpretation Introduction

Interpretation:Molesin $145.6\text{ L}$ of $\text{0}\text{.850 M}$ sodium cyanide should be calculated.

Concept introduction:The formula to evaluate moles from molarity is given as follows:

  $n=MV$

Where,

• $V$represents volume.
• $M$represents molarity.
• $n$represents number of moles.

The formula to convert mass in gram to moles is as follows:

  $\text{Mass}=\left(\text{Number of moles}\right)\left(\text{Molar mass}\right)$