Chapter 1, Problem 1.CE

(a)

Interpretation Introduction

Interpretation:

Concentration of $N{O}_3^{-}$ present in solution made of $12.6\text{ ppm}$ of dissolved $Ca{\left(N{O}_3\right)}_2$ has to be calculated in in ppm.

Concept introduction:

Parts Per Million refers to a term that expresses concentration of certain substance in a mixture of substances.   It is denoted as ppm.  It is represented as,

$ppm\text{ = }\frac{mass\text{ of substance}}{mass\text{ of sample}}\times {10}^6$

In solution concentrations, $1\text{ ppm}$ is equivalent to $1\mu \text{g/mL or 1 mg/L}\text{.}$

Molarity is a concentration term that is used to express the concentration of the solute in solution.  It is represented as,

$molarity\text{ = }\frac{no.of\text{ moles of solute}}{volume\text{ of solution in L}}$

Answer to Problem 1.CE

Concentration of $N{O}_3^{-}$ is calculated to be $9.52ppm.$

Explanation of Solution

Each mol of $Ca{\left(N{O}_3\right)}_2$ furnishes one mol $C{a}^{2+}$ ion and two mol $N{O}_3^{-}$ ions.  Formula of $N{O}_3^{-}$ $62.005$.  So the mass of Nitrate ion in $Ca{\left(N{O}_3\right)}_2$ is,

$\frac{2{\text{ mol NO}}_3^{-}}{1mol\text{ Ca}{\left(N{O}_3\right)}_2}\times \frac{62.005{\text{ g NO}}_3^{-}/mol{\text{ NO}}_3^{-}}{164.88\text{ g Ca}{\left(N{O}_3\right)}_2/mol\text{ Ca}{\left(N{O}_3\right)}_2}=0.7558{\text{ g NO}}_3^{-}/g\text{ Ca}{\left(N{O}_3\right)}_2$

Given that the solution made of $12.6\text{ ppm}$ of dissolved $Ca{\left(N{O}_3\right)}_2$ which implies the solution contains $12.6\mu \text{g of }Ca{\left(N{O}_3\right)}_2$ per $mL$ of solution or $\text{12}\text{.6 mg}$ of $Ca{\left(N{O}_3\right)}_2$ per liter or per gram of the solution.  $0.7558{\text{ g NO}}_3^{-}$ is present in per gram of $Ca{\left(N{O}_3\right)}_2$.  Thus $1\mu \text{g Ca}{\left({\text{NO}}_3\right)}_2$ contains $0.7558\mu {\text{g NO}}_3^{-}$ and so $0.7558\times \text{12}\text{.6 ppm = 9}\text{.52 ppm}$ of  $N{O}_3^{-}$ ions present in solution.

(b)

Interpretation Introduction

Interpretation:

The number of ppm of $Ca{\left(N{O}_3\right)}_2$ in $\text{0}\text{.144 m}M$ $Ca{\left(N{O}_3\right)}_2$ has to be calculated.

Concept introduction:

Parts Per Million refers to a term that expresses concentration of certain substance in a mixture of substances.   It is denoted as ppm.  It is represented as,

$ppm\text{ = }\frac{mass\text{ of substance}}{mass\text{ of sample}}\times {10}^6$

In solution concentrations, $1\text{ ppm}$ is equivalent to $1\mu \text{g/mL or 1 mg/L}\text{.}$

Molarity is a concentration term that is used to express the concentration of the solute in solution.  It is represented as,

$molarity\text{ = }\frac{no.of\text{ moles of solute}}{volume\text{ of solution in L}}$

Answer to Problem 1.CE

The number of ppm of $Ca{\left(N{O}_3\right)}_2$ in $\text{0}\text{.144 m}M$ $Ca{\left(N{O}_3\right)}_2$ is calculated to be $23.6ppm.$

Explanation of Solution

Given that concentration of $Ca{\left(N{O}_3\right)}_2$ in solution is $\text{0}\text{.144 m}M$ which is $\text{0}\text{.144 }\times {\text{10}}^{-3}M$.

As we know,

$molarity\text{ = }\frac{no.of\text{ moles of solute}}{volume\text{ of solution in L}}$

No. of moles of the solute $Ca{\left(N{O}_3\right)}_2$ is,

$no.of\text{ moles of }Ca{\left(N{O}_3\right)}_2\text{ = }\frac{\text{weight of }Ca{\left(N{O}_3\right)}_2}{\text{molecular weight of }Ca{\left(N{O}_3\right)}_2}$

Therefore molarity of $Ca{\left(N{O}_3\right)}_2$ is,

$molarity\text{ of }Ca{\left(N{O}_3\right)}_2\text{ = }\frac{\frac{mass\text{ of }Ca{\left(N{O}_3\right)}_2}{molar\text{ mass of }Ca{\left(N{O}_3\right)}_2}}{volume\text{ of solution in L}}$

Formula weight of $Ca{\left(N{O}_3\right)}_2$ is $164.088$

Let the volume of solution be $1\text{ L}$ and weight of $Ca{\left(N{O}_3\right)}_2$ required to make $\text{0}\text{.144 }\times {\text{10}}^{-3}M$ solution be ‘x’.  then,

No. of moles of $Ca{\left(N{O}_3\right)}_2$ is,

$\begin{array}{l}molarity\text{ of }Ca{\left(N{O}_3\right)}_2\text{= }\frac{\frac{x}{\text{164}\text{.088 g/mol}}}{\text{1 L}}\\ 0.144\times {10}^{-3}M=\frac{\frac{x}{\text{164}\text{.088 g/mol}}}{\text{1 L}}\\ \text{x}\text{= }0.144\times {10}^{-3}mol/L\times \text{164}\text{.088 g/mol}\\ =\text{ 23}\text{.6}\times {\text{10}}^{-3}\text{ g/L}\end{array}$

As we know, $1\text{ ppm = 1 mg/L}$,

$\begin{array}{l}23.6\times {10}^{-3}\text{ g/L}\text{= 23}\text{.6 }\times {\text{10}}^{-3}\times {10}^3\text{ mg/L}\\ \text{= 23}\text{.6 mg/L}\\ \text{= 23}\text{.6 ppm}\end{array}$

(c)

Interpretation Introduction

Interpretation:

The number of ppm of $N{O}_3^{-}$ in $\text{0}\text{.144 m}M$ $Ca{\left(N{O}_3\right)}_2$ has to be calculated.

Concept introduction:

Parts Per Million refers to a term that expresses concentration of certain substance in a mixture of substances.   It is denoted as ppm.  It is represented as,

$ppm\text{ = }\frac{mass\text{ of substance}}{mass\text{ of sample}}\times {10}^6$

In solution concentrations, $1\text{ ppm}$ is equivalent to $1\mu \text{g/mL or 1 mg/L}\text{.}$

Molarity is a concentration term that is used to express the concentration of the solute in solution.  It is represented as,

$molarity\text{ = }\frac{no.of\text{ moles of solute}}{volume\text{ of solution in L}}$

Answer to Problem 1.CE

The number of ppm of $N{O}_3^{-}$ in $\text{0}\text{.144 m}M$ $Ca{\left(N{O}_3\right)}_2$ is calculated to be $17.9ppm.$

Explanation of Solution

Given that concentration of $Ca{\left(N{O}_3\right)}_2$ in solution is $\text{0}\text{.144 m}M$ which is $\text{0}\text{.144 }\times {\text{10}}^{-3}M$. In previous step we found mass fraction of $N{O}_3^{-}$ in $Ca{\left(N{O}_3\right)}_2$ is $0.7558$ and $\text{0}\text{.144 m}M$ $Ca{\left(N{O}_3\right)}_2$ contains $23.6\text{ ppm}$ of $Ca{\left(N{O}_3\right)}_2$.  Therefore ppm of $N{O}_3^{-}$ in $\text{0}\text{.144 m}M$ $Ca{\left(N{O}_3\right)}_2$ is, $0.7558\times 23.6\text{ ppm = 17}\text{.9 ppm}$