Question

hardness of water is 20ppm the normality of caco3 in the water is in the form y×10^-3 then y is?​

Answer 1

Given:

• Hardness of water = 20 ppm
• Density of water = 1 g/mL
• Molar mass of CaCO₃ (Mm) = 100 g/mol

To find:

The value of y such that the normality of CaCO₃ is y*10⁻³.

Solution:

• One ppm hardness means that one gram of CaCO₃  is present every 10⁶g of water.
• Hence, 20 g of CaCO₃ (w) is present in 10⁶ g of water.
• Since, density of water = 1 g/mL, volume of solution (v) = 1000 L
• The n-factor of CaCO₃ (n) = 2
• Normality = (w*n)/(Mm*v) = 0.4*10⁻³

Answer:

The value of y such that the normality of CaCO₃ is y*10⁻³ = 0.4