Chapter 12, Problem 102A

(a)

Interpretation Introduction

Interpretation: The number of moles present in $47.8{\text{ g KNO}}_3$ is to be calculated.

Concept Introduction: One mole of a substance is the same in grams as one atom or molecule does in atomic mass units.

Answer to Problem 102A

The number of moles in $47.8{\text{ g KNO}}_3$ is $0.4728\text{ mol}$ .

Explanation of Solution

The given compound is $47.8{\text{ g KNO}}_3$ .

The molar mass of Potassium is $39.0983{\text{ g mol}}^{-1}$

The molar mass of Nitrogen is $14.0067{\text{ g mol}}^{-1}$

The molar mass of oxygen is $15.999{\text{ g mol}}^{-1}$

The molar mass of ${\text{KNO}}_3$ is calculated as follows:

  $\begin{array}{c}{\text{Molar mass KNO}}_3=\text{Molar mass K+Molar mass N+3}\times \text{Molar mass O}\\ =39.0983{\text{ g mol}}^{-1}+14.0067{\text{ g mol}}^{-1}+3\times 15.999{\text{ g mol}}^{-1}\\ =53.105{\text{ g mol}}^{-1}+47.997{\text{ g mol}}^{-1}\\ =101.102{\text{ g mol}}^{-1}\end{array}$

Therefore ${\text{1 mol KNO}}_3$ contains $101.102{\text{ g KNO}}_3$ .

The conversion factor to convert mass into moles is as follows:

  ${\text{g KNO}}_3\to {\text{mol KNO}}_3$

Convert $47.8{\text{ g KNO}}_3$ into moles as follows:

  $\text{47}\text{.8 }\bcancel{{\text{g KNO}}_3}\times \frac{{\text{1 mol KNO}}_3}{101.102\bcancel{{\text{g KNO}}_3}}=0.4728{\text{ mol KNO}}_3$

There, $0.4728\text{ mol}$ present in $47.8{\text{ g KNO}}_3$ .

(b)

Interpretation Introduction

Interpretation: The number of moles present in $2.22{\text{ L SO}}_2$ (at STP) is to be calculated.

Concept Introduction: Abbreviation Standard Temperature and Pressure is abbreviated as "STP." At STP, one mole of any gas will occupy a volume of $22.4\text{ L}$ .

Answer to Problem 102A

The number of moles in $2.22{\text{ L SO}}_2$ (at STP) is $0.0991\text{ mol}$ .

Explanation of Solution

The volume of ${\text{SO}}_2$ at STP is $2.22\text{ L}$ .

One mole of ${\text{SO}}_2$ is occupied $22.4\text{ L}$ at STP.

The conversion factor to convert volume into moles is as follows:

  ${\text{L SO}}_2\to {\text{mol SO}}_2$

Convert $2.22{\text{ L SO}}_2$ into moles as follows:

  $\text{2}\text{.22 }\bcancel{{\text{L SO}}_2}\times \frac{1{\text{ mol SO}}_2}{22.4\bcancel{{\text{L SO}}_2}}=0.0991{\text{ mol SO}}_2$

Therefore, $0.0991{\text{ mol SO}}_2$ is present in $2.22{\text{ L SO}}_2$ (at STP).

(c)

Interpretation Introduction

Interpretation: The number of moles present in $2.25\times {10}^{22}{\text{ molecules PCl}}_3$ (at STP) is to be calculated.

Concept Introduction: One mole of any substance contains $6.022\times {10}^{23}$ particles.

Answer to Problem 102A

The number of moles in $2.25\times {10}^{22}{\text{ molecules PCl}}_3$ is $0.0374\text{ mol}$ .

Explanation of Solution

The given number of molecules of ${\text{PCl}}_3$ is $2.25\times {10}^{22}\text{ molecules}$ .

One mole of ${\text{PCl}}_3$ containing $6.022\times {10}^{23}\text{ molecules}$ .

The conversion factor to convert molecules into moles is as follows:

  ${\text{molecules PCl}}_3\to {\text{mol PCl}}_3$

Convert $2.25\times {10}^{22}{\text{ molecules PCl}}_3$ into moles as follows:

  $2.25\times {10}^{22}\bcancel{{\text{molecules PCl}}_3}\times \frac{1{\text{ mol PCl}}_3}{6.022\times {10}^{23}\bcancel{{\text{molecules PCl}}_3}}=0.0374{\text{ mol PCl}}_3$

Therefore, $0.0374{\text{ mol PCl}}_3$ is present in $2.25\times {10}^{22}{\text{ molecules PCl}}_3$ .