Question

A sample of mixture of $CaC{l_2}$ and $NaCl$ weighing $4.22g$ was treated to precipitate all $Ca$ as $CaC{O_3}$ . This $CaC{O_3}$ is then heated and quantitatively converted into $0.959g$ of $CaO$. Calculate the percentage of $CaC{l_2}$ in the mixture $(Ca = 40,O = 16,C = 12,Cl = 35.5)$
A.$31.5\% $
B.$21.5\% $
C.$45.04\% $
D.$68.48\% $

Answer 1

Hint:It is important to determine the composition of a solution to understand the extent and feasibility of the solution and the composition of a solution is determined by the concentration of the solution which can be calculated by both quantitative and qualitative analysis. Mass percentage is a quantitative method to calculate the concentration of the solution.
Formula used:
$mass\% = \dfrac{\text{mass component}}{\text{mass of mixture}} \times 100$

Complete step by step answer:
The composition of a solution is important in determining the extent and feasibility of a solution. Composition of a solution can be determined by the concentration of a solution. Concentration can be analyzed quantitatively and qualitatively. There are several ways to determine quantitative analysis like mass percentage, volume percentage and various others.
Mass percentage is defined as the ratio of mass of the component in the solution to the total mass of the solution multiplied by 100. If a solution is said to contain $x\% $ of a component then this means that $x\,g$ of the component is present in $100\,g$ of solution.
For the above question we know that,
$CaC{O_3} \to CaO + C{O_2}$
Molecular mass of $CaC{O_3} = 100g/mol$
Molecular mass of $CaO = 56g/mol$
Thus, $100g$ of $CaC{O_3}$ gives $56g$ of $CaO$
Thus, $0.959gCaO = \dfrac{{0.959 \times 100}}{{56}}$
$ \Rightarrow 1.72gCaC{O_3}$
Molecular mass of $CaC{l_2} = 111g/mol$
Thus, $1.72gCaC{O_3} = \dfrac{{1.72 \times 111}}{{100}}$
$ \Rightarrow 1.9gCaC{l_2}$
So, the percentage of $CaC{l_2}$ in the mixture is given as,
$\% ofCaC{l_2} = \dfrac{\text{mass of } CaC{l_2}}{\text{mass of mixture}} \times 100$
$\% ofCaC{l_2} = \dfrac{{1.9}}{{4.22}} \times 100$
$ \Rightarrow 45.04\% $.

Thus, the percentage of $CaC{l_2}$ in the mixture is $45.04\% $ . So, the correct option is (C).

Note:
 Calcium is an important mineral which helps to build strong bones. Calcium carbonate is used as a supplement of calcium in our diet. It is used in production of cement, glass, steel and rubber. It is a salt of strong base and weak acid. Excessive calcium carbonate causes stomach pain and constipation and vomiting.