Question

A geochemist in the field takes a 46.0 mL sample of water from a rock pool lined with crystals of a certain mineral compound X. He notes the temperature of the pool, 21°C, and caps the sample carefully. Back in the lab, the geochemist filters the sample and then evaporates all the water under vacuum. Crystals of X are left behind. The researcher washes, dries and weighs the crystals. They weigh 0.87 g.
Required:
Using only the information above, can you calculate the solubility of X in water at 21°C? If yes, calculate it.

Answer 1

Answer: The solubility of X in water is $1.891 \times 10^{-2}$ g/ml.

Explanation:

The given data is as follows.

Volume of sample water = 46 ml

Temperature = $21^oC$

After vaporization, washes and then drying the weight of mineral X = 0.87 g

This means that 46.0 ml of water contains 0.87 g of X. Therefore, grams present in 1 ml of water will be calculated as follows.

          1 ml of water = $\frac{0.87 g}{46.0 ml}$

                                = $1.891 \times 10^{-2}$ g/ml

Therefore, we can conclude that solubility of X in water is $1.891 \times 10^{-2}$ g/ml.