Question

How do I calculate the number of chloride ions in a solution?
Which of the following contains the most chloride ions?
A. $10c{{m}^{3}}$ of $3.3\times {{10}^{-2}}mol\,d{{m}^{-3}}$ aluminium chloride solution
B. $20c{{m}^{3}}$ of $5.0\times {{10}^{-2}}mol\,d{{m}^{-3}}$ calcium chloride solution
C. $30c{{m}^{3}}$ of $3.3\times {{10}^{-2}}mol\,d{{m}^{-3}}$ hydrochloric acid
D. $40c{{m}^{3}}$ of $2.5\times {{10}^{-2}}mol\,d{{m}^{-3}}$ sodium chloride solution

Answer 1

In chemistry, mole is defined as the amount of the substance that contains $6.022\times {{10}^{23}}$ entities of a substance. This number is known as Avogadro’s number and it is denoted with the symbol ${{N}_{A}}$. Molarity is used for calculating the concentration of a solution. It is denoted with the symbol $M$.

Formula used:
$n=M\times V$
Where, $n$ is the number of moles, $M$ is the molarity and $V$ is the volume of the solution.

Complete step-by-step answer:
Let us discuss the given options one by one:
A. $10c{{m}^{3}}$ of $3.3\times {{10}^{-2}}mol\,d{{m}^{-3}}$ aluminium chloride solution
Here, the volume is equal to $10c{{m}^{3}}$ and the molarity of aluminium chloride solution is $3.3\times {{10}^{-2}}mol\,d{{m}^{-3}}$
Firstly, we will convert $c{{m}^{3}}$ to $d{{m}^{3}}$ using unitary method.
${{10}^{3}}c{{m}^{3}}$ is equal to $1d{{m}^{3}}$
$1c{{m}^{3}}$ is equal to $\dfrac{1d{{m}^{3}}}{{{10}^{3}}c{{m}^{3}}}$
$10c{{m}^{3}}$ is equal to $\dfrac{1d{{m}^{3}}}{{{10}^{3}}c{{m}^{3}}}\times 10c{{m}^{3}}$
$10c{{m}^{3}}$ is equal to ${{10}^{-2}}d{{m}^{3}}$
Now, to calculate moles, we will multiply molarity with volume of the solution.
$n=M\times V$
Where, $n$ is the number of moles, $M$ is the molarity and $V$ is the volume of the solution.
Now, we will substitute the values in the above formula, we get,
$n=\dfrac{3.3\times {{10}^{-2}}mol}{d{{m}^{3}}}\times {{10}^{-2}}d{{m}^{3}}$
$n=3.3\times {{10}^{-4}}mols$ of $AlC{{l}_{3}}$
B. $20c{{m}^{3}}$ of $5.0\times {{10}^{-2}}mol\,d{{m}^{-3}}$ calcium chloride solution
Here, the volume is equal to $20c{{m}^{3}}$ and the molarity of calcium chloride solution is $5.0\times {{10}^{-2}}mol\,d{{m}^{-3}}$
${{10}^{3}}c{{m}^{3}}$ is equal to $1d{{m}^{3}}$
$1c{{m}^{3}}$ is equal to $\dfrac{1d{{m}^{3}}}{{{10}^{3}}c{{m}^{3}}}$
$20c{{m}^{3}}$ is equal to $\dfrac{1d{{m}^{3}}}{{{10}^{3}}c{{m}^{3}}}\times 20c{{m}^{3}}$
$20c{{m}^{3}}$ is equal to $2\times {{10}^{-2}}d{{m}^{3}}$
Now, to calculate moles, we will multiply molarity with volume of the solution.
$n=M\times V$
Where, $n$ is the number of moles, $M$ is the molarity and $V$ is the volume of the solution.
Now, we will substitute the values in the above formula, we get,
$n=\dfrac{5.0\times {{10}^{-2}}mol}{d{{m}^{3}}}\times 2\times {{10}^{-2}}d{{m}^{3}}$
$n=1\times {{10}^{-3}}mols$ of $CaC{{l}_{2}}$
As $CaC{{l}_{2}}$ contains two chlorine atoms, so we will multiply the value of the number of moles with two. Hence, $n=2\times {{10}^{-3}}mols$ of $CaC{{l}_{2}}$
C. $30c{{m}^{3}}$ of $3.3\times {{10}^{-2}}mol\,d{{m}^{-3}}$ hydrochloric acid
Here, the volume is equal to $30c{{m}^{3}}$ and the molarity of hydrochloric acid is $5.0\times {{10}^{-2}}mol\,d{{m}^{-3}}$
${{10}^{3}}c{{m}^{3}}$ is equal to $1d{{m}^{3}}$
$1c{{m}^{3}}$ is equal to $\dfrac{1d{{m}^{3}}}{{{10}^{3}}c{{m}^{3}}}$
$30c{{m}^{3}}$ is equal to $\dfrac{1d{{m}^{3}}}{{{10}^{3}}c{{m}^{3}}}\times 30c{{m}^{3}}$
$30c{{m}^{3}}$ is equal to $3\times {{10}^{-2}}d{{m}^{3}}$
Now, to calculate moles, we will multiply molarity with volume of the solution.
$n=M\times V$
Where, $n$ is the number of moles, $M$ is the molarity and $V$ is the volume of the solution.
Now, we will substitute the values in the above formula, we get,
$n=\dfrac{3.3\times {{10}^{-2}}mol}{d{{m}^{3}}}\times 3\times {{10}^{-2}}d{{m}^{3}}$
$n=9.9\times {{10}^{-4}}moles$ of $HCl$
D.$40c{{m}^{3}}$ of $2.5\times {{10}^{-2}}mol\,d{{m}^{-3}}$ sodium chloride solution
Here, the volume is equal to $40c{{m}^{3}}$ and the molarity of sodium chloride solution is $2.5\times {{10}^{-2}}mol\,d{{m}^{-3}}$
${{10}^{3}}c{{m}^{3}}$ is equal to $1d{{m}^{3}}$
$1c{{m}^{3}}$ is equal to $\dfrac{1d{{m}^{3}}}{{{10}^{3}}c{{m}^{3}}}$
$40c{{m}^{3}}$ is equal to $\dfrac{1d{{m}^{3}}}{{{10}^{3}}c{{m}^{3}}}\times 40c{{m}^{3}}$
$40c{{m}^{3}}$ is equal to $4\times {{10}^{-2}}d{{m}^{3}}$
Now, to calculate moles, we will multiply molarity with volume of the solution.
$n=M\times V$
Where, $n$ is the number of moles, $M$ is the molarity and $V$ is the volume of the solution.
Now, we will substitute the values in the above formula, we get,
$n=\dfrac{2.5\times {{10}^{-2}}mol}{d{{m}^{3}}}\times 4\times {{10}^{-2}}d{{m}^{3}}$
$n={{10}^{-3}}moles$ of $NaCl$

Comparing all the values, we can observe that option (B) contains the maximum number of chloride ions. Therefore, the correct option is (B).

Note: In this question, we have concluded that molarity is the ratio of moles of solute to the volume of solution. In this question, we have calculated moles using this formula. One cubic decimeter is equal to ten cubic centimeter. Students can have a lot of problems in conversion of units. Hence, detailed knowledge is required to solve these kinds of questions.