Chapter 6, Problem 25A

(a)

Interpretation Introduction

Interpretation:

The mass in grams for the given sample has to be calculated.

Concept Introduction:

The formula to calculate the number of moles is given as shown below:

  $\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Answer to Problem 25A

The mass of carbon dioxide is $5.58\times {10}^{-2}\text{g}$ .

Explanation of Solution

The molar mass of carbon dioxide is $44.01\text{g}/\text{mol}$ . Substitute the value of given moles as $1.27\times {10}^{-3}$ and molar mass as $44.01\text{g}/\text{mol}$

in the above formula. The value of the mass of the given sample is calculated as shown below:

  $\begin{array}{c}1.27\times {10}^{-3}\text{ mol}=\frac{\text{Given mass}}{44.01\text{g}/\text{mol}}\\ \text{Given mass}=1.27\times {10}^{-3}\text{ mol}\times 44.01\text{g/mol}\\ =5.58\times {10}^{-2}\text{g}\end{array}$

(b)

Interpretation Introduction

Interpretation:

The mass in grams for the given sample has to be calculated.

Concept Introduction:

The formula to calculate the number of moles is given as shown below:

  $\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Answer to Problem 25A

The mass of nitrogen trichloride is $495.9\times {10}^3\text{g}$ .

Explanation of Solution

The molar mass of nitrogen trichloride is $120.365\text{g}/\text{mol}$ . Substitute the value of given moles as $4.12\times {10}^3$ and molar mass as $120.365\text{g}/\text{mol}$

in the above formula. The value of the mass of the given sample is calculated as shown below:

  $\begin{array}{c}4.12\times {10}^3\text{ mol}=\frac{\text{Given mass}}{120.365\text{g}/\text{mol}}\\ \text{Given mass}=4.12\times {10}^3\text{ mol}\times 120.365\text{g}/\text{mol}\\ =495.9\times {10}^3\text{g}\end{array}$

(c)

Interpretation Introduction

Interpretation:

The mass in grams for the given sample has to be calculated.

Concept Introduction:

The formula to calculate the number of moles is given as shown below:

  $\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Answer to Problem 25A

The mass of ammonium nitrate is $0.3609\text{g}$ .

Explanation of Solution

The molar mass of ammonium nitrate is $80.043\text{g}/\text{mol}$ . Substitute the value of given moles as $0.00451$ and molar mass as $80.043\text{g}/\text{mol}$

in the above formula. The value of the mass of the given sample is calculated as shown below:

  $\begin{array}{c}0.00451\text{mol}=\frac{\text{Given mass}}{80.043\text{g}/\text{mol}}\\ \text{Given mass}=0.00451\text{ mol}\times 80.043\text{g}/\text{mol}\\ =0.3609\text{g}\end{array}$

(d)

Interpretation Introduction

Interpretation:

The mass in grams for the given sample has to be calculated.

Concept Introduction:

The formula to calculate the number of moles is given as shown below:

  $\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Answer to Problem 25A

The mass of water is $324.288\text{g}$ .

Explanation of Solution

The molar mass of water is $18.016\text{g}/\text{mol}$ . Substitute the value of given moles as $18$ and molar mass as $18.016\text{g}/\text{mol}$

in the above formula and the value of the mass of the given sample is calculated as shown below:

  $\begin{array}{c}18\text{mol}=\frac{\text{Given mass}}{18.016\text{g}/\text{mol}}\\ \text{Given mass}=18\text{ mol}\times 18.016\text{g}/\text{mol}\\ =324.288\text{g}\end{array}$

(e)

Interpretation Introduction

Interpretation:

The mass in grams for the given sample has to be calculated.

Concept Introduction:

The formula to calculate the number of moles is given as shown below:

  $\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Answer to Problem 25A

The mass of copper (II) sulfate is $10007.4\text{g}$ .

Explanation of Solution

The molar mass of copper (II) sulfate is $159.609\text{g}/\text{mol}$ . Substitute the value of given moles as $62.7$ and molar mass as $159.609\text{g}/\text{mol}$

in the above formula. The value of the mass of the given sample is calculated as shown below:

  $\begin{array}{c}62.7\text{mol}=\frac{\text{Given mass}}{159.609\text{g}/\text{mol}}\\ \text{Given mass}=62.7\text{ mol}\times 159.609\text{g}/\text{mol}\\ =10007.4\text{g}\end{array}$