1 L of river water contains 6 mg `Mg^(2+)` and 20 mg `Ca^(2+)` ions as chloride salts. Determine the degree of hardness of that sample of river water.

Explanation

6 mg `Mg^(2+)` = 0.006 g `Mg^(2+)`
and20 mg `Ca^(2+)` = 0.02 g `Ca^(2+)` ion.
Now, 24 g `Mg^(2+) equiv ` 95 g `"MgCl"_(2) equiv 100 g " CaCO"_(3)`.
[ `because` atomic mass of Mg = 24 and molecular mass of `"MgCl"_(2)` = 95 and `"CaCO"_(3)` = 40 ]
`therefore` 0.006 g `Mg^(2+) = (100 xx 0.006)/(24)`g = 0.025 g `"CaCO"_(3)`.
Again, 40 g `Ca^(2+) equiv` 111 g `"CaCl"_(2)` = 100 g `"CaCO"_(3)` .
[`because` atomic mass of Ca = 40 & molecular mass of `"CaCl"_(2)` 111]
`therefore " "0.02 g " Ca"^(2+) = (100 xx 0.02)/(40)` g = 0.05 g `"CaCO"_(3)`.
So the quantity of `CaCO_(3)` equivalent to `"MgCl"_(2)` and
`CaCl_(2)` present in 1 L or `10^(3)` g water = (0.025 + 0.05) g = 0.075 g.
`thereofore " "10^(6)`g of water contains `(0.075 xx 10^(6))/(10^(3))` = 75 g of `CaCO_(3)`.
Hence, degree of hardness of the sample is 75 ppm.