A water sample contains 22.2mg of CaCl2 in 1 L of sample. The degree of hardness of water in terms of CaCO3 equivalent is ------ ppm. [. Molar mass of CaCO 3=100 g mol -1, molar mass of .CaCl 2=111 gmol -1]

Question

A water sample contains 22.2mg of CaCl2 in 1 L of sample. The degree of hardness of water in terms of CaCO3 equivalent is ------ ppm. [. Molar mass of CaCO 3=100 g mol -1, molar mass of .CaCl 2=111 gmol -1] (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

ppm is parts per million (106). This is equivalent to mg per litre. 1L of water contains 22.2mg of CaCl2 Now, 111g of CaCl2≡ 100gCaCO3 22.2mg of CaCl2=(22 . 2 × 100/111)=20 It is equivalent to 20mg text of CaCO3 text in 1L text of water ∴ Hardness of water sample =20ppm

Q. A water sample contains 22.2mg of CaCl2​ in 1 L of sample. The degree of hardness of water in terms of CaCO3​ equivalent is ------ ppm. [ Molar mass of CaCO3​=100gmol−1, molar mass of CaCl2​=111gmol−1]

ppm is parts per million (106). This is equivalent to mg per litre. 1L of water contains 22.2mg of CaCl2​ Now, 111g of CaCl2​≡100gCaCO3​ 22.2mg of CaCl2​=11122.2×100​=20 It is equivalent to 20mg of CaCO3​ in 1L of water ∴ Hardness of water sample =20ppm

Q. A water sample contains $22.2 m g$ of $C a C l_{2}$ in $1$ $L$ of sample. The degree of hardness of water in terms of $C a C O_{3}$ equivalent is ------ $pp m .$ $[$ Molar mass of $C a C O_{3} = 100 g m o l^{- 1}$ , molar mass of $C a C l_{2} = 111 g m o l^{- 1}]$