Question:
What is the degree of hardness in ppm of a sample containing \(19\text{ mg}\) of \(MgCl_2\) molecular weight \(=95\) in \(2\text{ kg}\) water sample? Express it in terms of equivalents of \(CaCO_3\).
What is the degree of hardness in ppm of a sample containing \(19\text{ mg}\) of \(MgCl_2\) molecular weight \(=95\) in \(2\text{ kg}\) water sample? Express it in terms of equivalents of \(CaCO_3\).
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Hardness as \(CaCO_3\) equivalent \(=\text{mass of salt}\times\frac{50}{\text{equivalent weight of salt}}\).
Updated On: May 7, 2026
- \(10\)
- \(20\)
- \(30\)
- \(40\)
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The Correct Option is A
Solution and Explanation
Given mass of \(MgCl_2\):
\[
19\text{ mg}.
\]
Molecular weight of \(MgCl_2\):
\[
95.
\]
Equivalent weight of \(MgCl_2\):
\[
\frac{95}{2}=47.5.
\]
Equivalent weight of \(CaCO_3\):
\[
50.
\]
Now convert \(MgCl_2\) into \(CaCO_3\) equivalent:
\[
\text{CaCO}_3\text{ equivalent}
=
19\times \frac{50}{47.5}.
\]
\[
=20\text{ mg}.
\]
This \(20\text{ mg}\) of \(CaCO_3\) equivalent is present in:
\[
2\text{ kg}
\]
of water.
Since:
\[
1\text{ kg water}\approx 1\text{ L water},
\]
we have:
\[
2\text{ kg water}\approx 2\text{ L water}.
\]
Hardness in ppm means mg of \(CaCO_3\) equivalent per litre.
So:
\[
\text{Hardness}=\frac{20}{2}.
\]
\[
=10\text{ ppm}.
\]
Hence, the degree of hardness is:
\[
10\text{ ppm}.
\]
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