Question

0g of a mixture of $BaO$ and $CaO$ requires 100 $cm^3$ of 2.5 M $HCl$ to react completely. The percentage of calcium oxide in the mixture is approximately:
(Given: molar mass of $BaO$=153)
A: 52.6
B: 55.1
C: 44.9
D: 47.4

Answer 1

The mole concept is very significant and useful in chemistry. It is actually the base of stoichiometry and it provides the best option to express the amounts of reactants as well as products that are consumed and formed during a chemical reaction.

Complete answer:
In the question, we are given a solution mixture of$BaO$ and $CaO$.
Mass of solution mixture = 10 g (Given)
Let us assume mass of $CaO$ as x g
Then, mass of $BaO$ = (10 – x) g
Now, we will find out the equivalent mass of $BaO$ and $CaO$.
We know that the equivalent weight of any element is equal to its atomic weight divided by the valence which it assumes in the compounds.
Therefore, equivalent mass of $BaO$ = ?
Molar mass of $BaO$=153 (Given)
$Equivalent{\text{ }}mass{\text{ }}of{\text{ }}BaO = \dfrac{{molar{\text{ }}mass}}{2} = \dfrac{{153}}{2} = 76.5$
And, to calculate the number of gram equivalents, we generally use the following formula:
$Number{\text{ }}of{\text{ }}gram{\text{ }}equivalents = \dfrac{{Given{\text{ }}mass{\text{ }}}}{{{\text{ }}Equivalent{\text{ }}mass{\text{ }}of{\text{ }}the{\text{ }}given{\text{ }}species}}$
$\therefore Number{\text{ }}of{\text{ }}gram{\text{ }}equivalents{\text{ }}of{\text{ }}BaO = \dfrac{{10 - x{\text{ }}}}{{{\text{76}}{\text{.5}}}}$
Similarly, we will calculate for $CaO$:
molar mass of $CaO$=56
$equivalent{\text{ }}mass{\text{ }}of{\text{ C}}aO = \dfrac{{molar{\text{ }}mass}}{2} = \dfrac{{56}}{2} = 28$
$\therefore Number{\text{ }}of{\text{ }}gram{\text{ }}equivalents{\text{ }}of{\text{ C}}aO = \dfrac{{x{\text{ }}}}{{28}}$
$\therefore Number{\text{ }}of{\text{ }}equivalents{\text{ }}of{\text{ }}HCl = 2.5 \times 0.1 = 0.25$ (Since concentration of $HCl$ = 2.5 M, volume = 100 $cm^3$ is given)
$Equivalents{\text{ }}of\;CaO\; + Equivalents{\text{ }}of\;BaO\; = \;Equivalents{\text{ }}of\;HCl$
Substituting the values, we will find out the value of x.

$$\Rightarrow \dfrac{x}{{28}} + \dfrac{{10 - x}}{{76.5}} = 0.25 \\\ \Rightarrow x = 5.26g$$

$\% {\text{ }}of{\text{ }}CaO = \dfrac{{5.26}}{{10}} \times 100 = 52.6\%$
The percentage of calcium oxide in the mixture is approximately $52.6\%$.

**Hence, the correct answer is Option A.

Note:**
- Never get confused between the use of normality, molarity and molality as a measure of concentration. Molarity is the number of moles of solute particles per litre of solution. Normality is the gram equivalent weight per litre of the solution. Molality is the number of moles of solute particles per kilogram of solvent.
- Most of the time, molarity is being preferred as the most suitable measure of concentration. In cases when the temperature of an experiment changes, then molality is preferred over other units of concentration. Normality is preferred most often in case of titration calculations.