Question

The concentration of Calcium ion in sea water is 0.0103 mol/kg. How many liters of seawater would you need to drink in order to have an intake of 1.03 grams of Ca (the RDA for adults is between 1-1.5 g/day)? The density of sea water is 1.020 g/mL.

Answer 1

Ans. Given,     [Ca²⁺] in sea water = 0.0103 mol/ kg

                        Density of sea water = 1.020 g/ mL = 1.020 kg/ L

Now,

            [Ca²⁺] in terms of “mol/L” = [Ca²⁺] x Density

                                                = (0.0103 mol/ kg) x (1.020 kg/ L)

                                                = 0.010506 mol/ L

So, there are 0.010506 mol Ca²⁺ ions per liter sea water.

# Mass of Ca²⁺ per liter sea water = moles of Ca²⁺ per L x Molar mass

                                                = (0.010506 mol/ L) x (40.078 g/ mol)

                                                = 0.421059468 g/ L

So, Ca²⁺ content in sea water is 0.4211 g/L

# Now,

            Required amount of sea water = Required Ca²⁺ intake / Ca²⁺ content of sea water

                                                = 1.03 g / (0.4211 g/ L)

                                                = 2.45 L

Therefore, required amount of sea water = 2.45 L