Question

The concentration of $Ca{{(HC{{O}_{3}})}_{2}}$ in a sample of hard water is 486 ppm. The density of the water sample is 1.0 g/ml. The molarity of the solution is:
(A) $3.0\times {{10}^{-3}}M$
(B) $5.0\times {{10}^{-3}}M$
(C) $2.0\times {{10}^{-3}}M$
(D) $6.0\times {{10}^{-3}}M$

Answer 1

Hint: To solve this question, we first need to find the "mg" concentration equivalent of "ppm" with the help of the density of water given. 1 ppm can be described as 1 part per million. It is usually used to depict contaminants.

Complete answer:
Since it is given to us that the density of water is 1g/ml, it means that 1 ml of water weighs 1 g or 0.001L of water weighs 1000mg.
\[1g/mL=\dfrac{1000mg}{0.001L}={{10}^{6}}mg/L\]
So, we can say that 1 liter of water weighs 1 million milligrams.
So, 1 milligram in 1 liter will be equivalent to 1 ppm.
Now, it is given to us that the concentration of $Ca{{(HC{{O}_{3}})}_{2}}$ in a sample of hard water is 486 ppm. This means that there is 486 mg or 0.486 g of $Ca{{(HC{{O}_{3}})}_{2}}$ in a 1-liter sample of hard water.
We know that the atomic mass of calcium is 40.078u, the atomic mass of hydrogen is 1.00794u, the atomic mass of carbon is 12.0107u, and the atomic mass of oxygen is 15.9994u.
So, the molecular molar mass of $Ca{{(HC{{O}_{3}})}_{2}}$ is
\[\begin{align}
  & {{M}_{Ca{{(HC{{O}_{3}})}_{2}}}}={{M}_{Ca}}+2\times [{{M}_{H}}+{{M}_{c}}+(3\times {{M}_{O}})] \\
 & {{M}_{Ca{{(HC{{O}_{3}})}_{2}}}}=40.078+2\times [1.00794+12.0107+(3\times 15.9994)] \\
 & {{M}_{Ca{{(HC{{O}_{3}})}_{2}}}}=162.11g/mol \\
\end{align}\]
Now, the molarity of a solution can be given by
\[M=\dfrac{{{w}_{s}}}{{{M}_{s}}}\times \dfrac{1000}{V}\]
Where ${{w}_{s}}$= mass of the solute dissolved (in grams),
${{M}_{s}}$= molar mass of the solute dissolved (in grams),
And V= total volume of the solution (in mL).
So, the molarity of the hard water solution will be
\[\begin{align}
  & M=\dfrac{0.486}{162.11}\times \dfrac{1000}{1000} \\
 & M\cong 3.0\times {{10}^{-3}}M \\
\end{align}\]

So, the correct answer is option (A) $3.0\times {{10}^{-3}}M$.

Note:
It should be noted that the SI unit for molarity which defines the concentration of a solute in a solution is in terms of the amount of solute per unit volume is mol/${{m}^{3}}$ but it is usually expressed in mol/$d{{m}^{3}}$ which is equivalent to mol/L. It can also be expressed by the capital letter M.