1.

Sea water contains 1272 g of Mg^(2+) per metric ton (1 mega gram). How much of slaked lime must be added to 1.0 metric ton of sea water to precipitate all the Mg^(2+) ion. (give approx value in kg).

Answer»

Solution :4 KG (approx)
`Mg^(2+) + Ca (OH)_(2) darr + Ca^(2+)`
74 gm
In sea water
24 gm
24 gm of `Mg^(2+) -= 74 G Ca(OH)_(2)`
`therefore""1272 g Mg^(@+) -= (74)/(24) xx 1272 = 3922 g`
`= 3.922 kg ~~ "4 kg of Ca"(OH)_(2)`


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