Question

a geochemist in the field takes a 14.0 ml sample of water from a rock pool lined with crystals of a certain mineral compound x. he notes the temperature of the pool, 29.°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 0.32 g. using only the information above, can you calculate the solubility of x in water at 29.°c? if you said yes, calculate it. be sure your answer has a unit symbol and the right number of significant digits.

Explanation:

Step1: Recall solubility formula

Solubility is defined as the mass of solute per volume of solvent. Here, the mass of the solute (compound X) is \(0.32\space g\) and the volume of the solvent (water) is \(14.0\space mL\).

Step2: Calculate solubility

The formula for solubility \(S=\frac{m}{V}\), where \(m\) is the mass of the solute and \(V\) is the volume of the solvent. Substituting \(m = 0.32\space g\) and \(V=14.0\space mL\) into the formula: \(S=\frac{0.32\space g}{14.0\space mL}\approx0.023\space g/mL\)

Answer:

yes, \(0.023\space g/mL\)