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 geochemi 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: Recall the definition of solubility

Solubility is defined as the amount of solute (in grams) that dissolves in 100 mL of solvent (water in this case) to form a saturated solution at a given temperature.

Step2: Set up the proportion

Let \(s\) be the solubility. We know that in \(17.0\) mL of water (original sample, since dilution doesn't change the amount of \(X\) dissolved in the original \(17.0\) mL sample), the amount of \(X\) is \(2.89\) g. Using the proportion \(\frac{s}{100\space mL}=\frac{2.89\space g}{17.0\space mL}\)

Step3: Solve for \(s\)

Cross - multiply: \(s=\frac{2.89\space g\times100\space mL}{17.0\space mL}\)

$$s=\frac{289}{17.0}\space g/100\space mL$$

$$s = 17.0\space g/100\space mL$$

Answer:

yes, \(17.0\space g/100\space mL\)