Question

a geochemist in the field takes a 17.0 ml sample of water from a rock pool lined with crystals of a certain mineral compound x. he notes the temperature of the pool, 15.°c, and caps the sample carefully. back in the lab, the geochemis first dilutes the sample with distilled water to 400. 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 2.89 g. using only the information above, can you calculate the solubility of x in water at 15.°c? if you said yes, calculate it. be sure your answer has a unit symbol and the right number of significant digits.

Explanation:

Step1: Determine the mass of compound \(X\) in the original sample

The mass of \(X\) obtained after evaporation is \(2.89\ g\). This mass was present in the diluted \(400\ mL\) sample. But we need to find the mass in the original \(17.0\ mL\) sample. Since dilution doesn't change the amount of solute, the mass of \(X\) in the original \(17.0\ mL\) sample is the same as the mass of \(X\) in the diluted \(400\ mL\) sample.

Step2: Calculate the solubility

Solubility is defined as the mass of solute per \(100\ mL\) of solvent.
Let \(s\) be the solubility. We use the proportion \(\frac{m}{V}=\frac{s}{100}\), where \(m = 2.89\ g\) (mass of solute) and \(V=17.0\ mL\) (volume of original solvent).

$$s=\frac{2.89\ g}{17.0\ mL}\times100\ mL$$

$$s=\frac{2.89\times100}{17.0}\ g$$

$$s = 17.0\ g$$

(rounded to three significant digits)

Answer:

yes, \(17.0\ g\)