Question

A chemistry student is given 600. mL of a clear aqueous solution at 37° 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 21° C. At that point, the student sees that a precipitate has formed. He pours off the remaining liquid solution, throws away the precipitate, and evaporates the water from the remaining liquid solution under vacuum. More precipitate forms. The student washes, dries and weighs the additional precipitate. It weighs 0.084 kg. Using only the information from above, can you calculate the solubility of X at 21° C?If yes, calculate it. Be sure your answer has a unit symbol and the right number of significant figures.

Answer 1

Answer:

The solubility is  $S = 0.0014 \ g$

Explanation:

From the question we are told that

   The volume of the solution is  $V = 600 mL$

    The initial temperature is  $T_i = 37 ^oC$

     The final temperature is $T_f = 21^oC$

      The additional precipitate is $m = 0.084 \ kg = 84 \ g$

Yes because the solubility of the substance X is the amount of X needed to saturate a unit volume of the solvent  (for solubility of a solute to be calculated the solute must be able to saturate the solvent)

now we see that the substance X saturated the solvent because a precipitate was formed which the student threw away

   The solubility at  21 ° C is mathematically represented as

            $S = \frac{m}{m_w * 100 g \ of water }$

Mass of water($m_w$) in the solution is mathematically represented as

       $m_w = V * \rho_w$

Where  $\rho = 1 \frac{g}{mL}$

So  

      $m_w =600 * 1$

       $m_w =600g$

So  

    $S = \frac{84}{600 * 100 g \ of water }$

    $S = 0.0014 \ g$