Question

If 0.5 g of CaCO3 is present in 1500 mL water, then the hardness of water (in ppm) is
[A] 333 [C] 256 [B] 500 • [D] 50​

Answer 1

$\huge\red{Answer}$

I think option b)

hope it helps ❤️

Answer 2

[A] 333

Explanation:

Given: mass of CaCO₃ = 0.5 g , water = 1500 g

Hardness of water in ppm = weight of CaCO₃/Weight of water × 10⁶

                                 = 0.5/1500 × 10⁶

                                 = 3.33  10⁻⁴ × 10⁶

                                 = 333 ppm