10L of hard water required 5.6g of lime for removing hardness. Hence temporary hardness in ppm of caco3.
The answer 1000
who helps me to solve it for me step by step thank a lot
To find the temporary hardness in ppm (parts per million) of CaCO3 using the given information, follow these steps:
Step 1: Convert the given quantity of lime to milligrams (mg) since ppm is commonly expressed in mg/L.
- 5.6g = 5600mg (1g = 1000mg)
Step 2: Calculate the concentration of lime in mg/L (ppm) using the formula:
- Concentration (ppm) = Mass of substance (mg) / Volume of solution (L)
Given:
- Mass of lime = 5600mg
- Volume of solution (hard water) = 10L
- Concentration (ppm) = 5600mg / 10L
- Concentration (ppm) = 560 mg/L
Step 3: Since temporary hardness is caused by the presence of calcium ions (Ca2+) and carbonate ions (CO3^-) in the water, we need to convert the concentration of lime to the concentration of CaCO3.
Calculation:
- The molecular weight of CaCO3 = (40.08 g/mol for Ca) + (12.01 g/mol for C) + (16.00 g/mol * 3 for O)
- Molecular weight of CaCO3 = 100.09 g/mol
The concentration of CaCO3 in ppm is equal to the concentration of lime (in ppm) multiplied by the ratio of the molecular weights of CaCO3 to lime (CaO).
- Concentration of CaCO3 (ppm) = Concentration of lime (ppm) * (Molecular weight of CaCO3 / Molecular weight of lime)
- Concentration of CaCO3 (ppm) = 560 mg/L * (100.09 g/mol / 56 g/mol)
Simplifying the equation:
- Concentration of CaCO3 (ppm) = 560 mg/L * 1.79
- Concentration of CaCO3 (ppm) ≈ 1000 ppm
Therefore, the temporary hardness in ppm of CaCO3 is approximately 1000 ppm.