Chapter 1, Problem 1.34P

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Dilution formula refers to a formula to obtain the quantity of concentrated solution to be withdrawn to make dilute solution of desired quantity.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

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

(b)

Interpretation Introduction

Interpretation:

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

Concept Introduction:

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,

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