Question

a chemistry student is given 300. ml of a clear aqueous solution at 24.°c. he is told an unknown amount of a certain compound x is dissolved in the solution. the student allows the solution to cool to 24.°c. the solution remains clear. he then evaporates all of the water under vacuum. a precipitate remains. the student washes, dries and weighs the precipitate. it weighs 0.036 kg. using only the information above, can you calculate the solubility of x in water at 24.°c? yes no if you said yes, calculate it. be sure your answer has a unit symbol and the right number of significant digits.

Explanation:

Step1: Convert mass units

Convert \(0.036\space kg\) to \(g\). Since \(1\space kg = 1000\space g\), then \(m = 0.036\times1000 = 36\space g\).

Step2: Assume water density

Assume the density of water is \(1\space g/mL\). So, the mass of \(300\space mL\) water is \(m_{water}=300\times1 = 300\space g\).

Step3: Calculate solubility

Solubility is defined as the mass of solute per \(100\space g\) of solvent. Let \(S\) be the solubility. Using the formula \(S=\frac{m_{solute}}{m_{water}}\times100\). Substitute \(m_{solute} = 36\space g\) and \(m_{water}=300\space g\) into the formula: \(S=\frac{36}{300}\times 100\space g\) per \(100\space g\) water.

Answer:

yes, \(12\space g\) per \(100\space g\) water