Chapter 1, Problem 1.BE

(a)

Interpretation Introduction

Interpretation:

Formal concentration of $48.0\text{ wt\%}$ solution of $HBr$ has to be calculated.

Concept Introduction:

Molarity, also formal concentration is one of the parameters used to express concentration of a solution.  It is expressed as,

$\text{Molarity = }\frac{number\text{ of }molesof\text{ solute}}{volume\text{ of solution in L}}$

Mass percent is one of parameters used to express concentration of a solution.  It is expressed as,

$\text{Mass percent of a solute = }\frac{mass\text{ of solute}}{mass\text{ of solution}}\times \text{ 100\%}$

Answer to Problem 1.BE

Formal concentration of $48.0\text{ wt\%}$ solution of $HBr$ is calculated as $8.90M.$

Explanation of Solution

Assume that volume of solution is $1.000\text{ L}$ which is equivalent to $1.50\text{ kg}$ as density of $HBr$ solution is $\text{1}\text{.50 g/mL}$. Then mass of $HBr$ used to prepare $48.0\text{ wt\%}$ solution $HBr$ is,

$\text{Mass percent of HBr = }\frac{mass\text{ of HBr}}{mass\text{ of solution}}\times \text{ 100\%}$

Rearranging the above expression,

$\text{mass of HBr = }Mass\text{ percent }\times \text{ mass of solution}$

Substitute the known values,

$\begin{array}{l}\text{mass of HBr }\text{= 48 }\%\times \text{ 1}\text{.50 kg }\\ \text{= 0}\text{.48}\times \text{ 1}\text{.50 kg }\\ =\text{ 0}\text{.72 kg HBr}\end{array}$

As we know,

$\begin{array}{l}Mass\text{ of solution = mass of solute + mass of solvent}\\ \text{ = mass of HBr + mass of water}\end{array}$

Mass of the solute, $HBr$ is calculated above. Therefore,

$\begin{array}{l}\text{1}\text{.50 kg of solution }\text{= 0}{\text{.72 kg of HNO}}_3\text{ + mass of water}\\ \text{mass of water }\text{= 1}\text{.50 kg - 0}\text{.72 kg }\\ \text{= 0}\text{.78 kg}\end{array}$

Number of moles of Hydrogen bromide is,

$\begin{array}{l}no.of\text{ moles of HBr}\text{= }\frac{mass\text{ of HBr}}{molar\text{ mass of HBr}}\\ =\frac{\text{720 g}}{\text{80}\text{.91 g/mol}}\text{ = 8}\text{.90 mol}\end{array}$

Molarity or formal concentration of the solution is calculated as,

$\begin{array}{l}\text{Molarity of HBr}\text{= }\frac{number\text{ of }moles\text{ HBr}}{volume\text{ of solution in L}}\\ =\frac{\text{8}\text{.90 mol}}{1.000\text{ L}}\\ \text{= 8}\text{.90 M}\end{array}$

(b)

Interpretation Introduction

Interpretation:

Mass of solution that contains $36g$ $HBr$ has to be calculated.

Concept Introduction:

Mass percent is one of parameters used to express concentration of a solution.  It is expressed as,

$\text{Mass percent of a solute = }\frac{mass\text{ of solute}}{mass\text{ of solution}}\times \text{ 100\%}$

Answer to Problem 1.BE

Mass of solution that contains $36g$ $HBr$ is calculated as $75g.$

Explanation of Solution

As we know,

$\text{Mass percent of HBr = }\frac{mass\text{ of HBr}}{mass\text{ of HBr solution}}\times \text{ 100\%}$

Mass percent of $HBr$ is given as $48.0\text{ wt\%}$ and plug it in above equation and rearrange it.

$\begin{array}{l}\text{Mass percent of HBr}\text{= }\frac{mass\text{ of HBr}}{mass\text{ of HBr solution}}\times \text{ 100}\\ \text{48}\text{.0}\text{= }\frac{36\text{ g}}{mass\text{ of HBr solution}}\times \text{100}\\ mass\text{ of HBr solution}=\frac{36\text{ g}}{\text{48}\text{.0}}\times 100\text{ = 75}\text{.0 g}\end{array}$

(c)

Interpretation Introduction

Interpretation:

Volume of solution that contains $233mmol\text{ HBr}$ has to be calculated.

Answer to Problem 1.BE

Volume of solution that contains $233mmol\text{ HBr}$ is calculated as $26.2mL.$

Explanation of Solution

In the previous step we calculated molarity of $HBr$ solution to be $8.90M.$ volume of solution is,

$\begin{array}{l}\text{Molarity of HBr}\text{= }\frac{number\text{ of }moles\text{ HBr}}{volume\text{ of solution in L}}\\ volume\text{ of solution in L}=\frac{\text{0}\text{.233 mol}}{8.90\text{ mol/L}}\\ \text{= 0}\text{.0262 L = 26}\text{.2 mL}\end{array}$

(d)

Interpretation Introduction

Interpretation:

The volume of solution required to prepare $0.250L$ of $0.160M$ $HBr$ has to be calculated.

Concept Introduction:

Dilution formula refers to a formula to obtain the quantity of concentrated solution to be withdrawn to make dilute solution of desired quantity.  The formula is given as,

$M_{conc}.V_{conc}=M_{dil}.V_{dil}$

Where,

$\begin{array}{l}{M}_{conc}=\text{ molar concentration of concentrated solution}\\ V_{conc}=\text{ volume of concentrated solution}\\ M_{dil}=\text{ molar concentration of diluted solution}\\ V_{dil}=\text{ volume of diluted solution}\end{array}$

Answer to Problem 1.BE

The volume of solution required to prepare $0.250L$ of $0.160M$ $HBr$ is calculated as $4.49mL.$

Explanation of Solution

Applying the dilution formula calculate the quantity of $8.90\text{M}$ $HBr$ that has to be diluted to $0.250L$ solution to make $0.160\text{ M}$ $HBr$.

The Dilution formula is given as,

$M_{conc}.V_{conc}=M_{dil}.V_{dil}\mathrm{......}\left(1\right)$

Let ‘x’ be the quantity of concentrated $HBr$ that has to be diluted to $0.250L$ dilute solution.  Thus,

$\begin{array}{l}{M}_{conc}=\text{ 8}\text{.90 M}\\ V_{conc}=\text{ x}\\ M_{dil}=\text{ 0}\text{.160 M}\\ V_{dil}=\text{ 0}\text{.250 L}\end{array}$

Plugging the above values in the equation (1),

$\begin{array}{l}8.90\text{ M }\times x=0.160\text{ M }\times 0.250\text{ L}\\ \text{ x}=\frac{\text{0}\text{.160 M }\times 0.250\text{ L}}{\text{8}\text{.90 M}}\\ =\text{ 0}\text{.00449 L}\\ \text{= 4}\text{.49 mL}\end{array}$