Question:medium

What is the degree of hardness in ppm of a sample containing \(19\) mg of \(MgCl_2\) \((\text{Molecular Weight}=95)\) in \(2\) kg water sample? Express it in terms of equivalents of \(CaCO_3\).

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Hardness is commonly expressed as \(CaCO_3\) equivalent in ppm.
  • \(10\)
  • \(20\)
  • \(30\)
  • \(40\)
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The Correct Option is A

Solution and Explanation

Step 1: Understanding the Concept:
The hardness of water is a measure of the concentration of dissolved multivalent cations (primarily Ca$^{2+}$ and Mg$^{2+}$). It is conventionally expressed as the equivalent concentration of calcium carbonate (CaCO$_3$) in parts per million (ppm). 1 ppm is equivalent to 1 mg of CaCO$_3$ per liter of water.
Step 2: Key Formula or Approach:
1. Find the number of moles of the hardness-causing salt (MgCl$_2$). 2. Determine the number of moles of CaCO$_3$ that are chemically equivalent to the moles of MgCl$_2$. The equivalence is based on the charge, but since both Ca$^{2+}$ and Mg$^{2+}$ have a +2 charge, 1 mole of MgCl$_2$ is equivalent to 1 mole of CaCO$_3$. 3. Calculate the mass of the equivalent CaCO$_3$. Mass = moles $\times$ Molar mass of CaCO$_3$. 4. Calculate the hardness in ppm. ppm = $\frac{\text{mass of CaCO}_3 \text{ equivalent (in mg)}}{\text{Volume of water (in L)}}$. (Assuming density of water is 1 kg/L).
Step 3: Detailed Explanation:
Given: - Mass of MgCl$_2$ = 19 mg = 0.019 g. - Molar mass of MgCl$_2$ = 95 g/mol. - Mass of water = 2 kg. Assuming the density of water is 1 kg/L, the volume of water is 2 L. - Molar mass of CaCO$_3$ = 40 (Ca) + 12 (C) + 3*16 (O) = 100 g/mol. 1. Calculate moles of MgCl$_2$: \[ \text{moles of MgCl}_2 = \frac{\text{mass}}{\text{molar mass}} = \frac{0.019 \text{ g}}{95 \text{ g/mol}} = 0.0002 \text{ mol} \] 2. Find equivalent moles of CaCO$_3$: The reaction equivalence is MgCl$_2 \equiv$ CaCO$_3$. So, moles of CaCO$_3$ equivalent = 0.0002 mol. 3. Calculate mass of equivalent CaCO$_3$: \[ \text{mass of CaCO}_3 = \text{moles} \times \text{molar mass} = 0.0002 \text{ mol} \times 100 \text{ g/mol} = 0.02 \text{ g} \] Convert this mass to milligrams: \[ 0.02 \text{ g} = 20 \text{ mg} \] 4. Calculate hardness in ppm: \[ \text{Hardness (ppm)} = \frac{\text{mass of CaCO}_3 \text{ equivalent (in mg)}}{\text{Volume of water (in L)}} \] \[ \text{Hardness (ppm)} = \frac{20 \text{ mg}}{2 \text{ L}} = 10 \text{ mg/L} = 10 \text{ ppm} \] Step 4: Final Answer:
The degree of hardness of the water sample is 10 ppm. Therefore, option (A) is correct.
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