Chapter 9, Problem 121QRT

(a)

Interpretation Introduction

Interpretation:

The fraction of chloride content in sample $1$ and sample $2$ has to be calculated and whether they can be distinguished by means of chemical analysis has to be determined.

Concept Introduction:

Chemical analysis is of two types: qualitative and quantitative.  It gives the stoichiometry of constituent elements in a mixture or compound which basically depends upon the physical and chemical properties.

Explanation of Solution

Sample $1$ has been prepared by grinding one mole of $KCl$ and one mole of $RbCl$ together .  Sample $2$ has been prepared by heating the mixture of one mole of $KCl$ and one mole of $RbCl.$  In both the sample the stoichiometry of chlorine is same.

Molar mass of $KCl$ and $RbCl$ are $74.55g/mol$ and $120.92g/mol$ respectively.

Mass of one mole of $KCl$ and one mole of $RbCl$ together is given below.

  $74.55g+120.92g=195.47g.$

Then, mass of chlorine in one mole of $KCl$ and one mole of $RbCl$ together is given below.

  $2\times 35.45g=70.90g.$

Fraction of chlorine present in both the samples can be calculated as given below.

  $\frac{70.90g}{195.47g}=\mathrm{0.3627.}$

Therefore, the fraction of chloride content in sample $1$ and sample $2$ is $0.3627.$

The samples can not be distinguished by means of chemical analysis.  Because both $KCl$ and $RbCl$ have same crystal structure and hence have same physical and chemical properties.

(b)

Interpretation Introduction

Interpretation:

The two X-ray diffraction results has to be interpreted in terms of the structures of the crystal lattices of sample $1$ and sample $2.$

Explanation of Solution

In case of sample $1$ , two different types of X-ray diffraction pattern has been observed – one with an edge length of $629pm$ and one with an edge length of $658pm.$  These unit cells are matching with the original crystal unit cells.  This indicates sample $1$ contains two separate salts.  That means two salts have been physically combined but not chemically combined.

In case of sample $2$ , a single diffraction pattern has been observed with an edge length of $640pm.$  The new unit cell is neither of the two original crystals.  This indicates that two salts have been chemically reorganized into a new crystal.

(c)

Interpretation Introduction

Interpretation:

The chemical formula for sample $1$ and sample $2$ has to be written.

Explanation of Solution

The sample $1$ is the mixture of one mole of $KCl$ and one mole of $RbCl$ salt and from the diffraction pattern it has decided that two slats have been physically combined but not chemically.  Hence, the chemical formula for sample $1$ can not be written.  For a mere understanding it can be written as $KCl+RbCl.$

From the X-ray diffraction studies, it has decided that two salts have been chemically reorganized into a new crystal.  Hence, the chemical formula for sample $2$ is $KRbC{l}_2.$

(d)

Interpretation Introduction

Interpretation:

The electrical conductivity of each $1.00g$ sample dissolved in $100mL$ of water has to be compared.  The name of the ion and its concentration in each solution has to be determined.

Answer to Problem 121QRT

The concentration of the ion $\left[K^{+}\right]$ , $\left[R{b}^{+}\right]$ and $\left[C{l}^{-}\right]$ are $0.0512M$ , $0.0512M$ and $0.102M$ respectively.

Explanation of Solution

Both the solution has been made by dissolving equal amount of each sample in $100mL$ of water.  So the volume is same and the mass of each substance is also same.  Suppose that both the samples ionize completely producing the same number of each kind of ions.  Hence, their electrical conductivities are also same.  Each sample gives $K^{+}$, $R{b}^{+}$ and $C{l}^{-}$ ions upon ionization.  Their concentration can be calculated as given below.

Molar mass of $KCl$ , $RbCl$ and the sample are $74.55g/mol$ , $120.92g/mol$ and $195.47g/mol$ respectively.

Consider the chemical formula $KRbC{l}_2.$

Then,

  $\begin{array}{l}\frac{1gsample}{100mL}\times \frac{1mol\left(KRbC{l}_2\right)}{195.47gsample}\times \frac{1mol{K}^{+}}{1mol\left(KRbC{l}_2\right)}\times \frac{1000mL}{1L}=0.0512M{K}^{+}\\ \\ \frac{1gsample}{100mL}\times \frac{1mol\left(KRbC{l}_2\right)}{195.47gsample}\times \frac{1mol{R}^{+}}{1mol\left(KRbC{l}_2\right)}\times \frac{1000mL}{1L}=0.0512MR{b}^{+}\\ \\ \frac{1gsample}{100mL}\times \frac{1mol\left(KRbC{l}_2\right)}{195.47gsample}\times \frac{2molC{l}^{-}}{1mol\left(KRbC{l}_2\right)}\times \frac{1000mL}{1L}=0.102MC{l}^{-}\end{array}$

Therefore, the concentration of the ion $\left[K^{+}\right]$ , $\left[R{b}^{+}\right]$ and $\left[C{l}^{-}\right]$ are $0.0512M$ , $0.0512M$ and $0.102M$ respectively.