Question

$ 50g $ of saturated aqueous solution of potassium chloride at $ {30^ \circ }C $ is evaporated to dryness. When $ 13.2g $ of dry $ KCl $ was obtained. The solubility of $ KCl $ in water at $ {30^ \circ }C $ is:
(A) $ 35.87g $
(B) $ 25.02g $
(C) $ 28.97g $
(D) $ 27.81g $

Answer 1

Hint :Provided solution is saturated aqueous solution of potassium chloride at 30 degrees. Water is evaporated and one can figure out the values of left-out mass comparable to the mass which has been evaporated and then calculate the final solubility.

Complete Step By Step Answer:
The solution containing $ KCl $ was evaporated by the supplication of heat and the remaining amount of $ KCl $ acquired is $ 13.2g $ .
First, we need to calculate the amount of evaporated water by subtracting it from total amount as follow:
  $ Mass{\text{ }}of{\text{ }}water = 50 - 13.2 = 36.8g $
So now we have the mass of evaporated water, we can now calculate the solubility by inserting the formula:
Solubility of $ KCl $ in water is mass of $ KCl $ obtained $ / $ mass of water evaporated $ \times $ $ 100g $
 $ \therefore So\operatorname{lub} ility\,of\,KCl\,in\,water = \dfrac{{13.2}}{{36.8}} \times 100 = 0.3586 \times 100 = 35.87g $
Hence, the solubility of $ KCl $ at $ {30^ \circ }C $ is $ 35.87g $ .
Additional Information: Temperature directly affects solubility. For most ionic solids, expanding the temperature builds how rapidly the solution can be made. Temperature can likewise expand the amount of solute that can be dissolved in a solvent.

Note :
The solubility of a substance relies upon the physical and chemical properties of the solute and solvent just as on temperature, pressure and presence of different chemicals. Solubility is the maximum concentration of a solute that can break down in a solvent at a given temperature. At the maximum concentration of solute, the solution is supposed to be soaked.