Suppose you wanted to prepare drinking water containing 100 mg/L of magnesium io

Question

Suppose you wanted to prepare drinking water containing 100 mg/L of magnesium ion (Mg^2+) by adding magnesium sulfate (MgSO_4) to previously purified water. Calculate the amount of MgSO_4 that you would have to add per liter to obtain 100 mg/L of Mg^2+. If you purchased a kilogram of reagent grade (99%) MgSO_4 at the current price of about $100/kg. how many liters of water having the calculated concentration of MgSO_4 could you prepare from a kilogram of MgSO_4? What would be the cost per liter of the added MgSO_4?

Explanation / Answer

2. To prepare a 100 mg/L Mg2+ water solution

1 mole of MgSO4 has 1 mole of Mg2+ in it

120.366 g of MgSO4 will have 24.305 g of Mg2+ in it

So, for 100 mg/L Mg2+ = 120.366 x 0.1/24.305

                                      = 0.495 g of MgSO4 is needed

Cost of 1 kg MgSO4 = $100/kg

Liters of water prepare with 100 mg/L Mg2+ concentration = 1000 g/0.1 g/L

                                                                                             = 10,000 Liters

Cost per liter of added MgSO4 = 10,000/100

                                                  = $10/L

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