when HgI2 is added to aqueous solution of KI why there is an increase in vapour pressure of solution ?
Answers
When HgI₂ reacts with the aqueous solution of KI , the following reaction takes place :
HgI₂ + 2 KI =====> K₂HgI₄
The number of molecules on the left hand side of the equation is 2 + 1 = 3 .
The number of molecules on the right hand side of the equation is 1 .
Hence there is a decrease in the number of molecules with the proceed of the reaction.
Thus the vapour pressure increases as the reaction proceeds .
Explanation :-
The product formed in the reaction is a complex salt .
Vapour pressure always decreases with the increase in solute and increases with the decrease in solute .
In the above problem , since the solute number goes down , the vapour pressure thus increases .
Freezing point of the solution will increase. Mercuric Iodide will react with Potassium Iodide to form dipotassium tetraiodomercury. The reaction is given below : 2KI + HgI2 = K2[HgI4] This association will decrease the number of ions in the solution as a result of which Van't Hoff Factor (i) will decrease as ndecreases from 4 to 3.