Question

a geochemist in the field takes a small sample of the crystals of mineral compound x from a rock pool lined with more crystals of x. he notes the temperature of the pool, 28.°c, and caps the sample carefully. back in the lab, the geochemist dissolves the crystals in 4.00 l of distilled water. he then filters this solution and evaporates all the water under vacuum. the researcher washes, dries and weighs the crystals. they weigh 0.960 kg. using only the information above, can you calculate the solubility of x in water at 28.°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 the formula for solubility

Solubility is defined as the amount of solute (in grams) per 100 mL (or 0.1 L) of solvent. Here, the mass of the solute \(X\) is \(m = 0.960\space kg=960\space g\) and the volume of water \(V = 4.00\space L\).

Step2: Calculate the solubility

The formula for solubility \(S\) is \(S=\frac{m}{V}\times0.1\). Substitute \(m = 960\space g\) and \(V = 4.00\space L\) into the formula.

$$S=\frac{960\space g}{4.00\space L}\times0.1=\frac{960\times0.1}{4.00}\space g/L = 24.0\space g/L$$

Answer:

yes, \(24.0\space g/L\)