A sample of hard water is analyzed and found to contain exactly of dissolved and exactly of dissolve — Hydrogen Chemistry Question
Question
A $1\text{ Litre}$ sample of hard water is analyzed and found to contain exactly $11.1\text{ mg}$ of dissolved $CaCl_2$ and exactly $12.0\text{ mg}$ of dissolved $MgSO_4$. What is the total degree of hardness of this water sample, expressed quantitatively in ppm (parts per million) of $CaCO_3$ equivalents? (Given Atomic masses: $Ca=40$, $Mg=24$, $Cl=35.5$, $S=32$, $C=12$, $O=16$)
Answer: 20
💡 Solution & Explanation
Molar masses: $CaCl_2 = 111$, $MgSO_4 = 120$, $CaCO_3 = 100$. Hardness from $CaCl_2$ in terms of $CaCO_3$ = $(100 / 111) \times 11.1\text{ mg} = 10\text{ mg}$. Hardness from $MgSO_4$ = $(100 / 120) \times 12.0\text{ mg} = 10\text{ mg}$. Total $CaCO_3$ equivalent = $10 + 10 = 20\text{ mg}$ per Litre, which equals $20\text{ ppm}$.
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