Question

A mixture of $CaC{l_2}$ and $NaCl$ weighing $4.44$ is treated with sodium to precipitate all the $C{a^{2 + }}$ ions as calcium carbonate. The calcium carbonate so obtained is heated strongly to get $0.56g$ of $CaO$. The percentage of $NaCl$ in the mixture of (atomic mass of $Ca = 40)$ is
A. 75
B. 30.6
C. 25
D. 69.4

Answer 1

Hint: As we know that solubility is the property of a solute to dissolve in a solvent. So, the percentage of a solution can be defined as the percent of a solute given a particular solution. The percent can be determined in two ways i.e. the ratio of the mass of the solute divided by the mass of the solution and the ratio of the volume of the solute divided by the volume of the solution.

Complete step by step answer:
The percentage of a solution by mass can be defined as the mass of a solute in the volume of a solution.
 Percentage of a solution (w/v) $ = \dfrac{{{\text{Mass of solute}}}}{{{\text{Volume of solution}}}} \times 100$
Percent by volume (v/v) $ = \dfrac{{{\text{Volume of solute}}}}{{{\text{Volume of solution}}}} \times 100$
According to the question, a mixture is formed when $CaC{l_2}$ and $NaCl$ reacts with sodium then it forms sodium carbonate. We can write the chemical equation for this reaction as,
$CaC{l_2} + NaCl + N{a_2}C{O_3} \to CaC{O_3} + 2NaCl + NaCl$
Both $CaC{l_2}$ and $NaCl$ weighing $4.44$ that is given in the question. The next statement is that calcium carbonate is obtained by heating strongly to get $0.56g$ of $CaO$. Then, the equation becomes:
$CaC{O_3}\xrightarrow{\Delta }CaO + C{O_2}$
$ 1 \;\;\;\;\;\;\;\;\;\;\;\;\;\;0.56g$
$100CaC{O_3} = 111gCaC{l_2}$
$ \Rightarrow 1gCaC{O_3} = 1.11gCaC{l_2}$
Thus, weight of $CaC{l_2}$= 1.11g
Then, we can calculate the weight of $NaCl$$ = \left( {4.44 - 1.11} \right) = 3.3g$
Now we can calculate percentage of NaCl as,
${\text{Percentage of solution}} = \dfrac{{{\text{Mass of solute}}}}{{{\text{Volume of solution}}}} \times 100$
Substituting the known values in the above formula we get,
Therefore, the percent of $NaCl$ = $\dfrac{{3.33}}{{4.44}} \times 100 = 75\% $

So, the correct answer is Option A.

Note: We must remember that the concentration of a solute is important for studying chemical reactions because it determines how often the molecules collide in a solution. Thus, concentration indirectly determines the rates of reactions and the conditions occurring at equilibrium. Also, it can be expressed in three different ways like percent by volume, percent by mass, and molarity.