
To $10{\text{ mL}}$ of $1{\text{ M}}$ $BaC{l_2}$ solution $5{\text{ mL}}$ of ${\text{0}} \cdot {\text{5 M}}$ ${K_2}S{O_4}$ is added. $BaS{O_4}$ is precipitated out. What will happen?
A. Freezing point will increase
B. Boiling point will increase
C. Freezing point will lower down
D. Boiling point will lower down
Answer
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Hint: The question is related to the effect of formed precipitate on the factors given in the option. One needs to understand the reaction and effect of $BaS{O_4}$ to the total solution as it is the precipitate formed in the solution. One can relate the concepts of freezing point and boiling point with the formed solution. This question may have multiple answers which are correct.
Complete step by step answer:
-First of all let's analyse the given reaction where the compound $BaS{O_4}$ is formed as precipitate. The precipitate $BaS{O_4}$ is a non-volatile compound which means it doesn’t evaporate from the formed solution and remains in the solution.
- The presence of non-volatile compounds in a solution lowers the vapour pressure of solution as it only acquires the surface area in the solution and doesn’t take part in vapourisation. So, when we increase the temperature of the solution by boiling it the solvent present in the solution starst to get vapourise but the non-volatile solute i.e. $BaS{O_4}$ will remain in solution increasing the density of solution. This will result in a more boiling point of solution as it needs more temperature to boil the solution. Hence, as the total boiling point of the solution will increase.
- Now lets see the effect of $BaS{O_4}$ precipitate on the freezing point of molecules which will get lowered as the precipitate formed will freeze at lower temperature than the pure form of solution earlier.
Therefore, the $BaS{O_4}$ precipitate formed in the solution will increase boiling point and lower down freezing point which shows the options B and C as the correct choice.
Note: The terms increase in boiling point and lowering in freezing point of solution are both colligative properties where these properties depend on the ratio between the number of solute particles present in the solution to the number of solvent molecules present in the solution. These properties do not depend on the nature of the chemical present and apply only to the dissolution of non-volatile solute in a volatile liquid solvent.
Complete step by step answer:
-First of all let's analyse the given reaction where the compound $BaS{O_4}$ is formed as precipitate. The precipitate $BaS{O_4}$ is a non-volatile compound which means it doesn’t evaporate from the formed solution and remains in the solution.
- The presence of non-volatile compounds in a solution lowers the vapour pressure of solution as it only acquires the surface area in the solution and doesn’t take part in vapourisation. So, when we increase the temperature of the solution by boiling it the solvent present in the solution starst to get vapourise but the non-volatile solute i.e. $BaS{O_4}$ will remain in solution increasing the density of solution. This will result in a more boiling point of solution as it needs more temperature to boil the solution. Hence, as the total boiling point of the solution will increase.
- Now lets see the effect of $BaS{O_4}$ precipitate on the freezing point of molecules which will get lowered as the precipitate formed will freeze at lower temperature than the pure form of solution earlier.
Therefore, the $BaS{O_4}$ precipitate formed in the solution will increase boiling point and lower down freezing point which shows the options B and C as the correct choice.
Note: The terms increase in boiling point and lowering in freezing point of solution are both colligative properties where these properties depend on the ratio between the number of solute particles present in the solution to the number of solvent molecules present in the solution. These properties do not depend on the nature of the chemical present and apply only to the dissolution of non-volatile solute in a volatile liquid solvent.
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