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100 ml sample is removed from a water...

100 ml sample is removed from a water solution with `CaSO_(4)` at `25^(@)C` . The water is completely evaorated from the sample and deposit of `0.24` gm `CaSO_(4)` is obtained . What is `K_(sp)` for `CaSO_(4)` at `25^(@)C`

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To find the solubility product constant (Ksp) for calcium sulfate (CaSO4) at 25°C, we can follow these steps: ### Step 1: Determine the mass of CaSO4 in 1 liter of solution Given that 0.24 grams of CaSO4 is obtained from a 100 ml sample, we can calculate how much would be present in 1 liter (1000 ml). \[ \text{Mass in 1 L} = \frac{0.24 \text{ g}}{100 \text{ ml}} \times 1000 \text{ ml} = 2.4 \text{ g} \] ...
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