10L of hard water required 5.6g of line for removing hardness. Hence temporary hardness in ppm of cac03.

The answer 1000
ppm= 1mg/1L
cac03 = 100 g/mol

who solve it for me clearly step by step

Thank a lot all tutors

To solve this problem, you need to follow these steps:

Step 1: Understand the Problem
The problem provides you with the information that 10L of hard water requires 5.6g of lime to remove the hardness. The objective is to find the temporary hardness in ppm (parts per million) of CaCO3.

Step 2: Calculate the Moles of CaCO3
The molar mass of CaCO3 is 100 g/mol. To calculate the number of moles, divide the given mass (5.6g) by the molar mass (100 g/mol):
Moles of CaCO3 = 5.6g / 100 g/mol = 0.056 mol

Step 3: Convert Moles to Mass in Mg
To convert moles to milligrams (mg), multiply the number of moles by 1000 (since 1 g = 1000 mg):
Mass of CaCO3 in mg = 0.056 mol * 1000 mg/mol = 56 mg

Step 4: Determine the Concentration in ppm
The concentration is measured in parts per million (ppm), which means milligrams per liter (mg/L). Since the problem states that 10L of hard water is used, the concentration in ppm is:
Concentration in ppm = 56 mg / 10 L = 5.6 ppm

Therefore, the temporary hardness in ppm of CaCO3 is 5.6 ppm.