Question

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.?
a. Yes
b. No
2) If yes than calculate the solubility of X. Round your answer to 3 significant digits

Answer 1

Step-by-step explanation:

1 ) The solubility can be calculated .

The sample taken from the pool was a saturated solution of mineral compound X because the pool was lined with the crystals of X .

The sample taken was 25 mL . Though it diluted afterwards , the solute contained in it did not change . The solute X was found to be 3.00 g

2 )

3 .00 g of solute X was contained in 25 mL of saturated solution .

solubility of X = 3 / 25 g per mL

= 3 x 1000 / 25 g per L

= 120 g / L

= 1.20 x 10² g / L