Chapter 6, Problem 21A

(a)

Interpretation Introduction

Interpretation : The number of moles present in 21.4 mg of nitrogen dioxide is to be calculated.

Concept Introduction :

  $Numberofmoles(n)=\frac{givenmassofsubs\tan ce}{molarmassofsubs\tan ce}$

Explanation of Solution

Given mass of nitrogen dioxide is: 21.4mg

The molar mass of nitrogen dioxide is: 46.0 g. /mol

Substitute these values in the above formula to get the number of moles.

  $\begin{array}{l}Numberofmoles(n)=\frac{givenmassofsubs\tan ce}{molarmassofsubs\tan ce}\\ n=\frac{21.4\times {10}^{-3}g}{46.0g/mol}\\ n=4.65\times {10}^{-4}mol\end{array}$

(b)

Interpretation Introduction

Interpretation: The number of moles present in 1.56g of copper nitrate is to be calculated.

Concept Introduction:

  $Numberofmoles(n)=\frac{givenmassofsubs\tan ce}{molarmassofsubs\tan ce}$

Explanation of Solution

Given mass of copper nitrate = 1.56g.

Molar mass of copper nitrate = 187.56 g./mol

  $\begin{array}{l}Numberofmoles(n)=\frac{givenmassofsubs\tan ce}{molarmassofsubs\tan ce}\\ n=\frac{1.56g}{187.56mol}\\ n=8.31\times {10}^{-3}mol\end{array}$

(c)

Interpretation Introduction

Interpretation: The number of moles present in 2.47 g of carbon disulfide to be calculated.

Concept Introduction:

  $Numberofmoles(n)=\frac{givenmassofsubs\tan ce}{molarmassofsubs\tan ce}$

Explanation of Solution

Given mass of carbon disulphide = 2.47g.

Molar mass of carbon disulphide = 76.1 g./mol

  $\begin{array}{l}Numberofmoles(n)=\frac{givenmassofsubs\tan ce}{molarmassofsubs\tan ce}\\ n=\frac{2.47g}{76.1g/mol}\\ n=3.24\times {10}^{-2}mol\end{array}$

(d)

Interpretation Introduction

Interpretation: The number of moles present in 5.04g of aluminum to be calculated.

Concept Introduction:

  $Numberofmoles(n)=\frac{givenmassofsubs\tan ce}{molarmassofsubs\tan ce}$

Explanation of Solution

Given mass of copper nitrate = 5.04 g.

Molar mass of copper nitrate = 342.1 g./mol

  $\begin{array}{l}Numberofmoles(n)=\frac{givenmassofsubs\tan ce}{molarmassofsubs\tan ce}\\ n=\frac{5.04g}{342.1g/mol}\\ n=1.47\times {10}^{-2}mol\end{array}$

(e)

Interpretation Introduction

Interpretation: The number of moles present in 1.56g of copper nitrate to be calculated.

Concept Introduction:

  $Numberofmoles(n)=\frac{givenmassofsubs\tan ce}{molarmassofsubs\tan ce}$

Explanation of Solution

Given mass of copper nitrate = 2.99g.

Molar mass of copper nitrate = 278.1 g./mol

  $\begin{array}{l}Numberofmoles(n)=\frac{givenmassofsubs\tan ce}{molarmassofsubs\tan ce}\\ n=\frac{2.99g}{278.1g/mol}\\ n=1.07\times {10}^{-2}mol\end{array}$

(f)

Interpretation Introduction

Interpretation: The number of moles present in 62.4g of calcium carbonate to be calculated.

Concept Introduction:

  $Numberofmoles(n)=\frac{givenmassofsubs\tan ce}{molarmassofsubs\tan ce}$

Explanation of Solution

Given mass of copper nitrate = 62.4g.

Molar mass of copper nitrate = 100.0 g./mol

  $\begin{array}{l}Numberofmoles(n)=\frac{givenmassofsubs\tan ce}{molarmassofsubs\tan ce}\\ n=\frac{62.4g}{100g/mol}\\ n=6.24\times {10}^{-1}mol\end{array}$