Question

The solubility of substance ‘X’ in water is $ 27.6 $ % (mass/volume) at $ 50 $ ℃. When $ 60 $ ml of saturated solution at $ 50 $ ℃. is cooled to $ 40 $ ℃, $ 3.7 $ grams of of solid substance ‘x’ is separated out the solubility of ‘x’ in water is:
(A) $ 30.7 $ %
(B) $ 16.56 $ % $ $
(C) $ 21.4 $ %
(D) $ 23.8 $ %

Answer 1

Hint: $ 50 $ %mass by volume means that $ 50 $ gm of solute is present in $ 100 $ ml of the solution thus formula used is:
 $ {\text{Solubility = }}\dfrac{{{\text{mass of solute(g)}}}}{{{\text{volume of solution(mL)}}}} \times 100 $

Complete step by step solution
Here at $ 50 $ ℃. we have
In a $ 100 $ ml solution $ 27.6 $ gm solute is present.
Thus, in $ 1 $ ml of solution $ 0.276 $ gm of solute is present.
In a $ 60 $ ml solution we have $ 0.276 $ $ \times 60 = 16.56 $ g of solute is present.
Now when the temperature is lowered to 40℃ then amount of X precipitated out is $ 3.7 $ g
Thus, amount of substance left in the solution is given by: $ 16.56 - 3.7 = 12.86 $ gm
The volume of the solution is still $ 60 $ ml thus
Solubility of X% is given by: $ {\text{Solubility = }}\dfrac{{{\text{mass of solute(g)}}}}{{{\text{volume of solution(mL)}}}} \times 100 $
Hence, we have $ = \dfrac{{12.86}}{{60}} = 0.214 $ g/ml
Thus, solubility in % is $ 0.214 \times 100 = 21.4\% $
Hence option C is the answer.

Note
Solubility can be defined in any term. It can be in terms of molality or molarity also. The process of approach remains the same.
We must keep in mind that 100ml of water is not the same as 100 g of water. Thus, the value of $ \dfrac{{{\text{mass}}}}{{{\text{mass}}}} $ % is often different from that of $ \dfrac{{{\text{mass}}}}{{{\text{volume}}}} $ %. In case of molality it is the solubility of the number of moles per kg of solvent and for molarity it is no. of moles per litre of solvent.