Q. Increasing amount of solid HgI2 is added to an aqueous solution containing 0.1 mol of KI. Which of the following graphs does represent the variation of depression in freezing point of the resulting solution with the amount of HgI2 added???
For all graphs, delta Tf is represented on Y-axis and moles of HgI2 added on X-axis:
a) A straight line inclined to the axis at an angle b/w 0o and 90o lying in the 1st quadrant having +ve y-intercept and -ve x-intercept, having no breaks and increasing as x increases.
b) A similar line as in a) but this time the line becomes parallel to x-axis beyond x = 0.05
c) A similar line as in a) but this time having a negaive slope and becomes parallel to x-axis beyon x = 0.05
d) A similar line as in c) but this time it becomes parallel to x-axis beyond x = 0.1.
See for the uploaded picture file if it gets uploaded. I hope u can imagine the graphs.... Plz tell me the answer somebody and along with reasoning...
Adding solute to solution decreases the freezing point of the solution because in a solution containing solute , solvent only freezes and solute remains intact. Decrease in freezing point is directly proportional to molality of the solution . We are adding more and more solute and some solvent has frozen ,therefore decrease in freezing point is increasing on increasing moles of HgI 2 . I think answer should be first option.
Hey chandni,
I think HgI2 at some stage will also form a complex with KI and so the answer may be b). Couz after that delta Tf will become constant.
I think HgI2 at some stage will also form a complex with KI and so the answer may be b). Couz after that delta Tf will become constant.
Both of you , i want to have a look at my view on the thermodynamics $G question and gimme your objections.
Ankul, you are correct. But then, what will happen when all the solvent has frozen and HgI2 forms a complex with KI. At that time, mass of solvent will be zero and delta T f would become infinity even when HgI 2 FORMS A COMPLEX WITH KI.