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Estimate the hardness of a sample of wat...

Estimate the hardness of a sample of water 1L of which contains 0.001 mol of dissolved `"MgCl"_(2)`

Text Solution

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We know, 1 mol `"MgCl"_(2) equiv 1 ` mol `CaCO_(3)`
or, 0.001 mol `"MgCl"_(2) equiv` 0.001 mol `"CaCO"_(3)`
Now, 0.001 mol `CaCO_(3) = 100 xx 0.001` g or 0.1 g of `"CaCO"_(3)`
So, in 1 L or 1000 g or `10^(3)` g water the amount of `CaCO_(3)`
equivalent to 0.001 mol `"MgCl"_(2)` is 0.1 g.
`therefore` In `10^(6)` g water, the amount of `"CaCO"_(3) " is"(0.1 xx 10^(6))/(10^(3))` g = 100 g
Hence, the degree of hardness of that sample of water = 100 ppm.
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