Question

A chemistry student adds a quantity of an unknown solid compound X to 750. mL of distilled water at 26.° C. After 10 minutes of stirring, all of the X has dissolved. The student then drains off the solution and evaporates the water under vacuum. A precipitate is left behind. The student washes, dries and weighs the precipitate. It weighs 0.012 kg.

Answer 1

Explanation:

The given data is as follows.

          X = 12 g,    Volume of water = 2 L,

Hence, solubility will be calculated as follows.

               $\frac{12 g}{2 L}$

              = 6 g/L

As we know that 1 L contains 1000 ml. Hence, the calculated solubility will be converted into g/ml as follows.

               Solubility = $\frac{6 g}{1000 ml}$

                               = 0.006 g/ml

Thus, we can conclude that the solubility of compound X in $26^{o}C$ is 0.006 g/ml.