Chapter 1, Problem 1.15P

(a)

Interpretation Introduction

Interpretation:

The volume of $\text{98 wt\%}$$H_{2}S{O}_4$ which has molarity $18.0M$, that has to be diluted to $\text{1}\text{.00 L}$ to make $\text{1}\text{.00 M}$$H_{2}S{O}_4$ has to be determined.

Concept Introduction:

Molarity, is one of the parameters used to express concentration of a solution.  It is defined as the number of moles of the solute dissolved per liter of the solution.  It is expressed as,

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

Weight percent is one of parameters used to express concentration of a solute and it is defined as percentage of weight of the solute present with respect to weight of the solution.  It is expressed as,

    $\text{Weight percent of a solute = }\frac{\text{weight of solute}}{\text{weight of solution}}\times \text{ 100\%}$

Dilution formula refers to a formula to obtain the quantity of concentrated solution to be withdrawn to make dilute solution of desired quantity.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 1.15P

The volume of $\text{98 wt\%}$$H_{2}S{O}_4$ which has molarity $18.0M$, that has to be diluted to $\text{1}\text{.00 L}$ to make $\text{1}\text{.00 M}$$H_{2}S{O}_4$ is calculated as $55.6mL.$

Explanation of Solution

Applying the dilution formula volume of $18.0M$$\text{98 wt\%}$$H_{2}S{O}_4$ that has to be withdrawn to make $\text{1}\text{.00 L}$ of $\text{1}\text{.00 M}$$H_{2}S{O}_4$

The Dilution formula is given as,

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

Let ‘x’ be the volume of $18.0M$ (concentrated solution) $\text{98 wt\%}$$H_{2}S{O}_4$ that has to be withdrawn for diluting to $1.00L$ of $\text{1}\text{.00 M}$$H_{2}S{O}_4$ (diluted)  solution.  Thus,

    $\begin{array}{l}{M}_{conc}=\text{ 18}\text{.0 M}\\ V_{conc}=\text{ x}\\ M_{dil}=\text{ 1}\text{.00 M}\\ V_{dil}=\text{ 1}\text{.00L}\end{array}$

Plugging the above values in the equation (1),

    $\begin{array}{l}18.0\text{ M }\times \text{ x}=\text{ 1}\text{.00 M }\times \text{ 1}\text{.00L}\\ \text{ x}=\frac{\text{1}\text{.00 M }\times \text{ 1}\text{.00 L}}{\text{18}\text{.0 M}}\\ =\text{ 0}\text{.0556 L = 55}\text{.6 mL}\end{array}$

(b)

Interpretation Introduction

Interpretation:

Density of $\text{98 wt\%}$$H_{2}S{O}_4$ has to be determined.

Concept Introduction:

Refer to part (a)

Answer to Problem 1.15P

Density of $\text{98 wt\%}$$H_{2}S{O}_4$ is calculated as $1.80g/mL.$

Explanation of Solution

Given that molarity of $\text{98 wt\%}$$H_{2}S{O}_4$ is $18.0M.$  The solution is diluted to volume of $1.00L$ and molarity $1.0M.$

Calculate the number of moles of $H_{2}S{O}_4$ in $\text{98 wt\%}$$H_{2}S{O}_4$

    $\begin{array}{l}moles\text{ of }{H}_{2}S{O}_4\text{= }Molarity\text{ of }{H}_{2}S{O}_4\times Volume\text{ of solution in L}\\ =18.0M\times 1.00L\\ =\text{ 18}\text{.0 mol}\end{array}$

Convert $\text{18}\text{.0 mol}$ to mass,

  $\begin{array}{l}\text{Mass of }{H}_{2}S{O}_4\text{= }Moles\text{ of }{H}_{2}S{O}_4\times Formula\text{ mass of }{H}_{2}S{O}_4\\ =\text{ 18 mol}\times 98.079\text{ g/mol = 1}\text{.77}\times {\text{10}}^3\text{ g }{H}_{2}S{O}_4\end{array}$

The solution contains $\text{98 wt\%}$$H_{2}S{O}_4$, mass of $H_{2}S{O}_4$ per milliliter is $\text{1}\text{.77g}$,

Density of solution is calculated as,

  $\text{density = }\frac{\text{mass}}{volume}$

Plug the above values to calculate density,

    $\begin{array}{l}\text{density = }\frac{1.77\text{ g}}{\text{ 0}\text{.980 g NaOH/ g solution}}\\ =\text{ 1}\text{.80 g/mL}\end{array}$