10L of hard water required 5.6g of line for removing hardness. hence temporary hardness in ppm of cac03. The answer is 1000
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To solve this problem step by step, we need to calculate the temporary hardness of the water in parts per million (ppm) of CaCO3. Here's how you can do it:
Step 1: Calculate the amount of CaCO3 required to remove the hardness.
- Given that 10L of hard water required 5.6g of lime (CaO), and considering the balanced chemical equation:
CaO + H2O → Ca(OH)2
We know that 1 mole of CaO reacts with 1 mole of Ca(OH)2, and the molar mass of Ca(OH)2 is approximately 74 g/mol.
- To calculate the number of moles of CaO:
Moles of CaO = Mass of CaO / Molar mass of CaO
= 5.6g / 56 g/mol
= 0.1 mol
Step 2: Calculate the moles of CaCO3.
- Since 1 mole of CaO reacts with 1 mole of CaCO3, the moles of CaCO3 will be the same as the moles of CaO:
Moles of CaCO3 = 0.1 mol
Step 3: Calculate the concentration of CaCO3 in ppm.
- The ppm value represents the number of parts of a substance per one million parts of water.
- One part per million is equal to one milligram per liter (mg/L) or one milligram per kilogram (mg/kg). To convert moles to milligrams, we need to use the molar mass of CaCO3, which is approximately 100 g/mol.
- To calculate the concentration of CaCO3 in ppm:
Concentration in ppm = Moles of CaCO3 * Molar mass of CaCO3 / Volume of water in liters
= 0.1 mol * 100 g/mol / 10 L
= 1 g/L
= 1000 ppm
Therefore, the temporary hardness of the water in ppm of CaCO3 is 1000.