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The hardness of a water sample containing 10^{-3}M MgSO_{4} expressed as CaCO_{3} equivalents (in ppm) is_____________.

(molar mass of MgSO_{4} is 120.37 g/mol)

 

Option: 1

100ppm


Option: 2

200ppm


Option: 3

99.85ppm


Option: 4

111.5ppm


Answers (1)

best_answer

We know that,

The hardness of Water : 
The hardness of water is due to the presence of bicarbonates, chlorides, and sulfates of calcium and magnesium bicarbonates causes temporary hardness while chlorides and sulphates cause permanent hardness. The extent of hardness is called degree of hardness. It is defined as the number of parts by weight of calcium carbonate present In one million parts by weight of water that is, in ppm (milligrams per litre) of CaCO3
 

\mathrm{Hardness\, of\, water\, =\, \frac{gm\, of\, calcium\, carbonate}{10^{6} gm\, of\, water}}

\\\mathrm{ppm_{(in\: terms\: of\: CaCO_{3})}\: =\: \frac{weight\ of\ CaCO_3\ in\ gram}{weight\ of\ water\ in\ gram}\: x\: 10^{6}} \\\textrm{1 L = 1000\ gram}

 
Now, Need to find the weight gram of CaCO3

10–3 molar CaSO4 = 10–3 moles of CaSO4 present in 1 L solution.

We assume 

nCaCO3 = nCaSO4 

weight of CaCO3 / molar mass of CaCO3  = mole of CaSO4

weight of CaCO3 /100  = 10-3

weight of CaCO3 = 100 X 10-3

And the volume of water = 1 L then 

weight of water = 1000 

\mathrm{ppm_{(in\: terms\: of\: CaCO_{3})}\: =\: \frac{10^{-3}\: x\: 100}{1000}\: x\: 10^{6}}

\mathrm{ppm_{(in\: terms\: of\: CaCO_{3})}\: =\: 100ppm}

Hence, the option number (1) is correct.

Posted by

shivangi.shekhar

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