Question

0.00012 % Mgso4, and 0.000 111% CaCl2 is present in water, what is the measured hardness of water a millimoles of washing soda respectively required to purify 1000 L of this water. (a) 1 ppm & 10 millimoles (c) 2 ppm & 20 millimoles (b) 2 ppm & 10 millimoles (d) 1 ppm & 20 millimoles

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Answer 1

Answer:

What is the question I did not understand the question

Answer 2

Answer:

given = 0.00012 % Mgso4,

            0.000 111% CaCl2

            1000 L water

to find = hardness of water

solution =

MgSO₄ = 0.00012g = 0.00012 / 120 mole

CaCl₂  = 0.000111g = 0.00111 / 111 mole

equvivalentmoles of CaCO₃ =  ( 0.00012 /120 + 0.000111 / 111) mole

mass of CaCO₃ =  0.00012/120 +  0.000111 /111 X  100 = 2 X 10⁻⁴

hardness of CaCO₃=  2 X 10⁻⁴ / 100 X 10⁸ = 2ppm

CaCl₂ + Na₂CO₃ → CaCO₃ + 2NaCl

NaSO₄ + Na₂CO₃ → Mgco3 + NaSO4

Na₂CO₃ for 100g of water = 2 X 10⁻⁴Mole

Na₂CO₃ for 1000 lit water = 2 X 10⁻⁶/100X10⁶= 2/100 mole

                                           =  20 mole

so the option C is the correct answer.