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).

Explanation

To determine how much slaked lime (calcium hydroxide, Ca(OH)₂) must be added to 1.0 metric ton of sea water to precipitate all the Mg²⁺ ions, we can follow these steps:

### Step 1: Understand the Reaction
The reaction between magnesium ions (Mg²⁺) and slaked lime (Ca(OH)₂) can be represented as:
\[ \text{Mg}^{2+} + \text{Ca(OH)}_2 \rightarrow \text{Mg(OH)}_2 \downarrow + \text{Ca}^{2+} \]

This indicates that one mole of Mg²⁺ reacts with one mole of Ca(OH)₂ to produce one mole of magnesium hydroxide (Mg(OH)₂), which precipitates out of solution.

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