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0 votes

When mercuric iodide is added to the aqueous solution of potassium iodide,the

$\begin{array}{1 1}(a)\;\text{freezing point is lowered}\\(b)\;\text{freezing point is raised}\\(c)\;\text{freezing point does not change}\\(d)\;\text{boiling point does not change}\end{array}$

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3 Answers

+1 vote

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. Now, Freezing point depression when a solute is added to a solvent is given by ΔT = i Kf m Clearly now i has decreased from when it was an aqueous KI solution to when HgI is added so ΔT is less now, hence less depression in freezing point which means a higher freezing point. Hence the freezing point of solution will increase. Note: Van't Hoff factor (i) =1+alpha(n-1) where alpha=dissociation degree and n is the no of ions dissociated Hope you understand the reasoning behind it now.

answered Dec 6, 2015 by vivekloya24
Vant hoff factor (i) depends on the number of particles(n). In the reactants you have KI and HgI2. KI would give 2 particles and HgI2 would give 3 particles. The total is 5 particles (and not 4 as suggested by you).Hence (i)decreases from 5 to 3
0 votes
answer is B

donit know the reason but it a previous year jee question(1987)
answered Dec 6, 2015 by vivekloya24
–2 votes
Depression in freezing point of solution is proportional to the molality of solution.
$\therefore$ By adding mercuric iodide,the molality of solution $(I^-)$ increases,hence freezing point is lowered.
Hence (a) is the correct answer.
answered Feb 14, 2014 by sreemathi.v
answer is B but as per previous year jee question(1987)
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