Chapter 1, Problem 111QRT

Interpretation Introduction

Interpretation:

From the given, the mass of silver in each sample has to be calculated.  Also, the mass of chlorine in the sample of $\text{AgCl}$ and the mass of iodine in the sample of $\text{AgI}$ has to be calculated.

Explanation of Solution

The mass of each atom to the mass of sample is given as

  $\begin{array}{l}\text{1}\text{.0000}\text{g}\text{AgCl}\text{=}{\text{m}}_{\text{Ag}}\text{+}{\text{m}}_{\text{Cl}}\\ \\ \text{1}\text{.6381}\text{g}\text{AgI}\text{=}{\text{m}}_{\text{Ag}}\text{+}{\text{m}}_{\text{I}}\\ \\ {\text{m}}_{\text{I}}\text{=}\text{3}\text{.580}{\text{m}}_{\text{Cl}}\end{array}$

Elimination of ${\text{m}}_{\text{Ag}}$ from the above expression:

  $\text{1}\text{.6381}\text{g}\text{-}\text{1}\text{.0000}\text{g}\text{=}{\text{m}}_{\text{Ag}}{\text{+m}}_{\text{I}}\text{-}{\text{(m}}_{\text{Ag}}\text{+}{\text{m}}_{\text{Cl}}\text{)}\text{=}{\text{m}}_{\text{I}}\text{-}{\text{m}}_{\text{Cl}}$

The value of ${\text{m}}_{\text{I}}$ is substituted to get ${\text{m}}_{\text{Cl}}$.

  $\begin{array}{l}\text{1}\text{.6381}\text{g}\text{-}\text{1}\text{.0000}\text{g}\text{=}{\text{m}}_{\text{I}}\text{-}{\text{m}}_{\text{Cl}}\\ \\ \text{0}\text{.6381}\text{g}\text{=}\text{3}\text{.580}{\text{m}}_{\text{Cl}}\text{-}{\text{m}}_{\text{Cl}}\text{=}\text{2}\text{.580}{\text{m}}_{\text{Cl}}\\ \\ {\text{m}}_{\text{Cl}}\text{=}\text{0}\text{.2473}\text{g}\text{Cl}\end{array}$

The ${\text{m}}_{\text{Ag}}$ is calculated as

  $\begin{array}{l}{\text{m}}_{\text{Ag}}\text{=}\text{1}\text{.0000}\text{g}\text{AgCl}\text{-}{\text{m}}_{\text{Cl}}\\ \\ \text{=}\text{1}\text{.0000}\text{g}\text{AgCl}\text{-}\text{0}\text{.2473}\text{g}\text{Cl}\\ \\ \text{=}\text{0}\text{.7527}\text{g}\text{Ag}\end{array}$

The ${\text{m}}_{\text{I}}$ is calculated as

  $\begin{array}{l}{\text{m}}_{\text{I}}\text{=}\text{1}\text{.6381}\text{g}\text{AgI}\text{-}{\text{m}}_{\text{Ag}}\\ \\ \text{=}\text{1}\text{.6381}\text{g}\text{AgI}\text{-}\text{0}\text{.7527}\text{g}\text{Ag}\\ \\ \text{=}\text{0}\text{.8854}\text{g}\text{I}\end{array}$

The ratio of calculated mass of Iodine to Chlorine is

  $\frac{\text{0}\text{.8854}\text{g}}{\text{0}\text{.2473}\text{g}}\text{=}\text{3}\text{.580}$

The mass of chlorine in $\text{AgCl}$ is $\text{0}\text{.2473}\text{g}$.

The mass of Iodine in $\text{AgI}$ is $\text{0}\text{.8854}\text{g}$.

The mass of silver in each sample is $\text{0}\text{.7527}\text{g}$.