Question

One liter of a sample water contains $4.44mg$ $CaC{l_2}$ and $1.9mg$ of $MgC{l_2}$, what is the total hardness in terms of ppm of $CaC{O_3}$
(A) $2$ ppm
(B) $3$ ppm
(C) $4$ ppm
(D) $6$ ppm

Answer 1

Hint:Hardness of water is defined as the amount of calcium and magnesium dissolved in water. This is expressed as an equivalent quantity of calcium carbonate. Calculate the amount of $CaC{O_3}$ produced by the given weights of $CaC{l_2}$ and $MgC{l_2}$ which will give the hardness of water in ppm.

Complete step-by-step solution:Given to us that one liter of water contains $4.44mg$ $CaC{l_2}$ and $1.9mg$ of $MgC{l_2}$.
We know that one mole of $CaC{l_2}$ and one mole of $MgC{l_2}$ are equivalent to one mole of $CaC{O_3}$
Now, let us calculate the molecular weights of each compound separately.
Molecular weight of $CaC{l_2}$ is $40 + 2 \times 35.5 = 111$ grams.
Molecular weight of $MgC{l_2}$ is $24 + 2 \times 35.5 = 95$ grams.
Similarly, the molecular weight of $CaC{O_3}$ is $40 + 12 + 3 \times 16 = 100$ grams.
Since, we know that the moles of each compound are equivalent, we can write that one mole of $CaC{O_3}$ produces one mole of $CaC{l_2}$
This means that $100$ grams of $CaC{O_3}$ produces $111$ grams of $CaC{l_2}$.
Let us assume that X mg of $CaC{O_3}$ is required to produce $4.44mg$ of $CaC{l_2}$
Hence, the value of X would be $X = \dfrac{{100 \times 4.44}}{{111}} = 4mg$
Similarly we can write for $MgC{l_2}$ as follows.
Now, $100$ grams of $CaC{O_3}$ produces $95$ grams of $MgC{l_2}$
Let us assume that Y mg of $CaC{O_3}$ is required to produce $1.9mg$ of $MgC{l_2}$
So, we can calculate the value of Y as $Y = \dfrac{{1.9 \times 100}}{{95}} = 2$ mg.
So the total amount of hardness present in water would be $4 + 2 = 6$ ppm.

Therefore, the correct answer is option (D).

Note: Note that $1$ mole $CaC{l_2} \equiv 1$ mole $MgC{l_2} \equiv 1$ mole $CaC{O_3}$. We know that the weight of one mole of any compound would be its molecular weight and hence we can derive the weights of Calcium carbonate using the above equivalence.