Chapter 7, Problem 7.73AP

(a)

Interpretation Introduction

Interpretation:

The grams of sodium nitrate present have to be calculated.

Concept Introduction:

Molarity:  Molarity is defined as the mass of solute in one liter of solution.  Molarity is the preferred concentration unit for stoichiometry calculations.  The formula is,

  $\text{Molarity=}\frac{\text{Mass}\text{of}\text{solute(in}\text{moles)}}{\text{Volume}\text{of}\text{solution}\text{(in}\text{litres)}}$

Answer to Problem 7.73AP

The grams of sodium nitrate is $\text{3}\text{.23g}\text{.}$

Explanation of Solution

Given,

Volume of solution=$\text{150mL}$

Molarity of solution=$\text{0}\text{.25M}$

Milliliters is converted into liters,

  $\begin{array}{l}\text{L=150mL×}\frac{\text{1L}}{\text{1000mL}}\\ \text{L=0}\text{.150L}\end{array}$

The moles of sodium nitrate is calculated as,

  $\begin{array}{l}\text{Molarity=}\frac{\text{Mass}\text{of}\text{solute(in}\text{moles)}}{\text{Volume}\text{of}\text{solution}\text{(in}\text{litres)}}\\ \text{0}\text{.25M=}\frac{\text{x}}{\text{0}\text{.150L}}\\ \text{x=}\left(\text{0}\text{.150L}\right)\left(0.25\frac{\text{mol}}{\text{L}}\right)\\ \text{x=0}\text{.0375mole}\end{array}$

The moles of sodium nitrate is $\text{0}\text{.038mol}\text{.}$

Moles to grams is converted as,

  $\begin{array}{l}\text{Grams=0}{\text{.038mol NaNO}}_{\text{3}}\text{×}\frac{\text{85}{\text{.00g NaNO}}_{\text{3}}}{\text{1mole}{\text{NaNO}}_{\text{3}}}\\ \text{Grams=3}\text{.23g}\end{array}$

The grams of sodium nitrate is $\text{3}\text{.23g}\text{.}$

(b)

Interpretation Introduction

Interpretation:

The grams of nitric acid present have to be calculated.

Concept Introduction:

Refer to part (a).

Answer to Problem 7.73AP

The grams of nitric acid is $\text{5}\text{.7g}\text{.}$

Explanation of Solution

Given,

Volume of solution=$\text{45mL}$

Molarity of solution=$\text{2}\text{.0M}$

Milliliters is converted into liters,

  $\begin{array}{l}\text{L=45mL×}\frac{\text{1L}}{\text{1000mL}}\\ \text{L=0}\text{.045L}\end{array}$

The moles of nitric acid is calculated as,

  $\begin{array}{l}\text{Molarity=}\frac{\text{Mass}\text{of}\text{solute(in}\text{moles)}}{\text{Volume}\text{of}\text{solution}\text{(in}\text{litres)}}\\ \text{2}\text{.0M=}\frac{\text{x}}{\text{0}\text{.045L}}\\ \text{x=}\left(\text{0}\text{.045L}\right)\left(2.0\frac{\text{mol}}{\text{L}}\right)\\ \text{x=0}\text{.09mole}\end{array}$

The moles of nitric acid is $\text{0}\text{.09mol}\text{.}$

Moles to grams is converted as,

  $\begin{array}{l}\text{Grams=0}{\text{.09mol HNO}}_3\text{×}\frac{\text{63}{\text{.02g HNO}}_3}{\text{1mole}{\text{HNO}}_3}\\ \text{Grams=5}\text{.6718g}\end{array}$

The grams of nitric acid is $\text{5}\text{.7g}\text{.}$

(c)

Interpretation Introduction

Interpretation:

The grams of hydrochloric acid present have to be calculated.

Concept Introduction:

Refer to part (a).

Answer to Problem 7.73AP

The grams of hydrochloric acid is $\text{138}\text{.548g}\text{.}$

Explanation of Solution

Given,

Volume of solution=$\text{2}\text{.5L}$

Molarity of solution=$\text{1}\text{.5M}$

The moles of hydrochloric acid is calculated as,

  $\begin{array}{l}\text{Molarity=}\frac{\text{Mass}\text{of}\text{solute(in}\text{moles)}}{\text{Volume}\text{of}\text{solution}\text{(in}\text{litres)}}\\ \text{1}\text{.5M=}\frac{\text{x}}{\text{2}\text{.5L}}\\ \text{x=}\left(\text{2}\text{.5L}\right)\left(\text{1}\text{.5}\frac{\text{mol}}{\text{L}}\right)\\ \text{x=3}\text{.75mole}\end{array}$

The moles of hydrochloric acid is $\text{3}\text{.8mol}\text{.}$

Moles to grams is converted as,

  $\begin{array}{l}\text{Grams=3}\text{.8mol HCl×}\frac{\text{36}\text{.46g HCl}}{\text{1mole}\text{HCl}}\\ \text{Grams=138}\text{.548g}\end{array}$

The grams of hydrochloric acid is $\text{138}\text{.548g}\text{.}$