Question

A sample of hard water is found to contain $50mg$ of $CaS{{O}_{4}}$ in $12Kg$ of the sample. Calculate the hardness of the sample in terms of ppm of $CaS{{O}_{4}}$.

Answer 1

Here, ppm stands for parts per million. We can say that if a sample is to be calculated in ppm, then it equals the mass of the solute in gram to that of mass of the solution in gram multiplied with the million times.

Complete answer:
Given that,
A sample of hard water is found to contain $50mg$ of $CaS{{O}_{4}}$ in $12Kg$ of the sample. We have to find out the hardness of the sample in terms of ppm of $CaS{{O}_{4}}$.
SO, as per the given question:
The mass of the solute ($CaS{{O}_{4}}$) given is $50mg$ which can also be mentioned as $0.05g$.
The mass of the solvent (water) given is $12Kg$, which can also be mentioned as $12000g$.
So, the total mass of the solution will be the sum of the mass of the solute and solvent.
Thus, the mass of the solution will be $(0.05+12000)g=12000.05g$.
We know, ppm which is referred to as the parts per million can be calculated as the mass of solute to that of the total mass of the solution a million times.
Hence, the hardness of the sample in terms of ppm of $CaS{{O}_{4}}$ can be calculated as:
$ppm=\dfrac{\text{Mass of solute(g)}}{\text{Mass of solution(g)}}\times {{10}^{6}}$
Here, so by placing the above values:
$Hardness=\dfrac{0.05}{12000.05}\times {{10}^{6}}=4.16ppm$
Hence, the hardness of the sample in terms of ppm of $CaS{{O}_{4}}$ will be $4.16ppm$.

Note:
There are two types of hardness of water that are permanent and temporary. Permanent hardness is present, if water contains calcium and magnesium metals chloride or carbonates while if calcium and magnesium hydrogen carbonates are present, then the hardness is temporary hardness.