A geochemist in the field takes a 36.0 mL sample of water from a rock pool lined with crystals of a certain mineral compound X. He notes the temperature of the pool, 170 C, and caps the sample carefully. Back in the lab, the geochemist first dilutes the sample with distilled water to 500. 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 3.96 g Using only the information above, can you calculate yes the solubility of X in water at 17.0 C? If you said yes, calculate it. Be sure your answer has a unit symbol and 3 O no significant digits.

The water of a rock pool lined with mineral crystals is a saturated solution of said mineral, this means the concentration of X in those 36 mL is the solubility of compound X in water at 17 °C.

This means it is possible to calculate said solubility.

The dilution of the sample is not relevant, nor is that 500 mL volume. What's important is that 3.96 g of X form a saturated solution with 36.0 mL of water, so the solubility is:

3.96 g / 36.0 mL = 0.110 g/mL