Chapter 10.3, Problem 46PP

a.

Interpretation Introduction

Interpretation:

The formula of silver chromate needs to be determined.

Concept introduction:

The molecular formula of silver chromate can be determined by using the formula of ions present in it. The number of ratio in which two ions are present should be such that total charge on the compound is zero.

Answer to Problem 46PP

The formula of silver chromate is ${\text{Ag}}_{\text{2}}{\text{CrO}}_{\text{4}}\text{.}$

Explanation of Solution

In silver chromate, cation and anion are as follows:

Cation: $A{g}^{+}$

Anion: $Cr{O}_4^{-2}$

Here, charge on Ag is +1 and that on chromate is -2. Thus, Ag and chromate combines in 2:1 ratio. The formula will be:

${\text{Ag}}_{\text{2}}{\text{CrO}}_{\text{4}}$

Interpretation Introduction

(b)

Interpretation:

The number of cations in the sample needs to be determined.

Concept introduction:

In 1 mol of any substance, there are $6.022\times {10}^{23}$ units of that substance. These units can be atoms and molecules. The number of units in 1 mol of any substance is known as Avogadro’s number. It is denoted by symbol N_A.

Answer to Problem 46PP

The number of cations present in the sample is $9.3\times {10}^{22}$ ions.

Explanation of Solution

Given information:

A sample of silver chromate has a mass of $25.8$ g.

Calculation:

The mass of the silver chromate and molar mass is 25.8 g and 331.73 g/mol respectively.

The number of moles can be calculated as follows:

$n=\frac{m}{M}=\frac{25.8\text{ g}}{331.73\text{ g/mol}}=0.078\text{ mol}$

From the formula 1 mol of compound contains 2 mol of cation or Ag⁺. Also, in 1 mol there are $6.022\times {10}^{23}$ ions thus, number of cations will be:

$\begin{array}{l}\text{Cations present=2×0}{\text{.077×N}}_{\text{A}}\\ =\text{2×0}\text{.077×6}\text{.022}\times {\text{10}}^{23}\\ \text{=9}{\text{.3×10}}^{\text{22}}\end{array}$

c.

Interpretation Introduction

Interpretation:

The number of anions in the sample needs to be determined.

Concept introduction:

In 1 mol of any substance, there are $6.022\times {10}^{23}$ units of that substance. These units can be atoms and molecules. The number of units in 1 mol of any substance is known as Avogadro’s number. It is denoted by symbol N_A.

Answer to Problem 46PP

The number of anions present in the sample is $4.64\times {10}^{22}$ ions

Explanation of Solution

Given information:

A sample of silver chromate has a mass of $25.8$ g.

Calculation:

The mass of the silver chromate and molar mass is 25.8 g and 331.73 g/mol respectively.

The number of moles can be calculated as follows:

$n=\frac{m}{M}=\frac{25.8\text{ g}}{331.73\text{ g/mol}}=0.078\text{ mol}$

From the formula 1 mol of compound contains 1 mol of anion or $Cr{O}_4^{-2}$. Also, in 1 mol there are $6.022\times {10}^{23}$ ions thus, number of cations will be:

$\begin{array}{l}\text{Cations present=1}\times \text{0}{\text{.077×N}}_{\text{A}}\\ =\text{1×0}\text{.077×6}\text{.022}\times {\text{10}}^{23}\\ \text{=4}{\text{.64×10}}^{\text{22}}\end{array}$

d.

Interpretation Introduction

Interpretation:

The mass of 1 formula unit of silver chromate needs to be determined.

Concept introduction:

In 1 mol of any substance, there are $6.022\times {10}^{23}$ units of that substance. These units can be atoms and molecules. The number of units in 1 mol of any substance is known as Avogadro’s number. It is denoted by symbol N_A.

Answer to Problem 46PP

The mass in grams of one formula unit of silver chromate is $5.512\times {10}^{-22}g.$

Explanation of Solution

The molar mass of silver chromate is 332 g/mol thus, mass of 1 mol is 332 g. In 1 mol there are $6.022\times {10}^{23}$ molecules. Thus, this is mass of $6.022\times {10}^{23}$ of silver chromate.

Now, mass of 1 molecule will be:

$m=\frac{M}{N_A}=\frac{332}{N_A}g=\frac{332}{6.022\times {10}^{23}}=5.512\times {10}^{-22}g$