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A geochemist in the field takes a 25.0 mL sample of water from a rock pool lined with crystals of a certain mineral compound X. He notes the temperature of the pool, 26.° C, and caps the sample carefully. Back in the lab, the geochemist first dilutes the sample with distilled water to 350. mL. Then he filters it and evaporates all the water under vacuum. Crystals of X are left behind. The researcher washes, dries and weighs the crystals. They weigh 3.00 g.1) Using only the information above, can you calculate the solubility of X in water at 26 degrees Celsius.? Yes or No2) If yes than calculate the solubility of X. Round your answer to 3 significant digits

Sagot :

Answer:

Yes we can calculate the solubility

The solubility of the mineral is 120 g/L

Explanation:

We have the information that the sample taken originally from the rock pool is 25.0 mL

So,

25.0 mL of the sample contained 3.000 g of the sample

10000mL of the sample now contains 1000 * 3.000/25 = 120 g

This means that 120 g of the sample dissolves in 1000mL or 1L of solution.

Therefore, the solubility of the mineral is 120 g/L

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