Equal volumes of $\pu{1.0 M}$ $\ce{KCl (aq)}$ and $\pu{1.0 M}$ $\ce{AgNO3 (aq)}$ solutions are mixed. The depression of freezing point of the resulting solution (with respect to water) will be (assume $K_{f}$ for water is $\pu{1.86 K * kg / mol}$ and molarity ≈ molality).
A) $\pu{1.86 K}$ B) $\pu{3.72 K}$ C) $\pu{0.93 K}$ D) $\pu{7.44 K}$

My answer is.... Since for freezing point the vapour pressure of solute and solution should be same thus adding the mixture would result in same vapour pressure as before... Thus the depression in freezing point should be equal to $\pu{1.86 K}$.

Am I right or wrong? Please help!

  • 4
    $\begingroup$ Please don't post questions twice! You can edit or delete your own questions. $\endgroup$
    – Karl
    Jan 23, 2019 at 18:47
  • 2
    $\begingroup$ For formatting, See here and here. For a more detailed MathJax guide, look here, minor other details $\endgroup$ Sep 11, 2020 at 11:01
  • $\begingroup$ The previous version was correct for multiple reasons.. do not revert a fixed system.. $\endgroup$ Sep 11, 2020 at 11:02

1 Answer 1


Vapour pressure is not involved in the reasoning.

Freezing-point depression is in fact a colligative property, so it depends on the total number of solute particles and not of their chemical nature, so as you say the freezing-point depression is the same for both solutions ($\ce{KCl}$ and $\ce{AgNO3}$).

However, note that the depression is proportional to the total number of particles - and in the case of strong electrolytes (such as both $\ce{KCl}$ and $\ce{AgNO3}$), we have twice the number of ions in solution than the formal concentration of the salt would indicate.

Given the expression for the freezing-point descent:

$$ \Delta T_{f} = K_{f} · \sum_{i} m_{i} $$

where $\Delta T_{f}$ is the cryoscopic descent, $K_{f}$ is the cryoscopic constant, and $m_{i}$ is the molality of dissolved species $i$.

Since we have strong electrolytes that we can assume are completely dissociated,

$$ m_{\ce{K+}} = m_{\ce{Cl-}} = M_{\ce{KCl}} $$ $$ m_{\ce{Ag+}} = m_{\ce{NO3-}} = M_{\ce{AgNO3}} $$ $$ m_{\ce{KCl}} = m_{\ce{AgNO3}} = 0 $$

where $M$ is the formal molality of the salt. Therefore:

$$ \Delta T_{f} = K_{f} \left( m_{\ce{K+}} + m_{\ce{Cl-}} + m_{\ce{Ag+}} + m_{\ce{NO3-}} \right) = 2 K_{f} \left( M_{\ce{KCl}} + M_{\ce{AgNO3}} \right) $$

Since we're mixing equal volumes (and disregarding any mixing volume change), the formal molalities of both salts in the final solution will be $\pu{0.5 mol / kg}$, and their sum $\pu{1.0 mol / kg}$:

$$ \Delta T_{f} = 2 · \pu{1.86 K * kg / mol} · \pu{1.0 mol / kg} = \pu{3.72 K} $$

So the correct answer would be B) (but see the note below).

Note that you can find an alternative formulation for the expression of the freezing-point depression, using the van't Hoff factor:

$$ \Delta T_{f} = K_{f} \sum_{j} M_{j} · i_{j} $$

where $i_{j}$ is the van't Hoff factor for a salt, i.e. the number of ions into which it dissociates in solution. Again, for our completely dissociated strong electrolytes, $i_{\ce{KCl}} = i_{\ce{AgNO3}} = 2$, which gives exactly the same expression as before.

Note on solubility: the presence of $\ce{Ag+}$ and $\ce{Cl-}$ will result in the formation of a solid precipitate, $\ce{AgCl (s)}$. This will in practice remove $\ce{Ag+}$ and $\ce{Cl-}$ ions from the solution; however, since no value for the solubility product $K_{\rm{s}}$ is given, and no mention is made of this precipitation in the question, I've assumed it is not considered. If in the context of your course or text this is something you'd have to take into account (i.e you are expected to recognise that $\ce{AgCl}$ is insoluble in water and would precipitate in this situation), since you are not provided with a value for $K_{\rm{s}}$, we can only assume that the precipitation is complete, which would remove all $\ce{Ag+}$ and $\ce{Cl-}$ ions from the solution:

$$ \ce{Ag+ + Cl- -> AgCl \downarrow} $$

$$ m_{\ce{Ag+}} = m_{\ce{Cl-}} \approx 0 $$

$$ \Delta T_{f} = K_{f} \left( m_{\ce{K+}} + m_{\ce{NO3-}} \right) \approx \pu{1.86 K} $$

Which makes A) the right answer.

  • $\begingroup$ Alright! Thank you bro! But a small question. The molality m(kcl) and m(AgNo3) should be equal to 1 and not zero? Right? And how is M1×i1 + M2×i2 = 2?? Shouldn't it be 4? $\endgroup$
    – Asad Ahmad
    Jan 23, 2019 at 12:26
  • $\begingroup$ 1) Since $\ce{KCl}$ and $\ce{AgNO3}$ are strong electrolytes, they undergo complete dissociation: $\ce{KCl -> K+ + Cl-}$, $\ce{AgNO3 -> Ag+ + NO3-}$, so all the salt is in the form of separate cations and anions, and there is no undissociated $\ce{KCl}$ or $\ce{AgNO3}$ "unit" in solution, hence why there is no "real" concentration of $\ce{KCl}$ or $\ce{AgNO3}$ and why the concentration of the separate ions is equal to the "formal" (theoretical) concentration of the salts. Does that make sense? $\endgroup$
    – user41033
    Jan 23, 2019 at 12:36
  • $\begingroup$ 2) On the second part, keep in mind that $\pu{1.0 mol/kg}$ is the molality of the salts prior to mixing. When you mix equal volumes, you are diluting the same amount of $\ce{KCl}$ and $\ce{AgNO3}$ in twice as much volume - the solvent coming from their own solution plus the same amount of solvent coming from the other solution. Hence, once we've mixed them, the formal molality ($M_{i}$) of both salts is not $\pu{1.0 mol/kg}$ but $\pu{0.5 mol/kg}$ - and therefore $\pu{0.5 mol/kg} · 2 + \pu{0.5 mol/kg} · 2 = \pu{2.0 mol/kg}$. $\endgroup$
    – user41033
    Jan 23, 2019 at 12:40
  • 1
    $\begingroup$ Any teacher should expect that half of his students recognise the insoluble combination of silver and choride ions. OK, a teacher who gives out multiple-choice tests ... $\endgroup$
    – Karl
    Jan 23, 2019 at 18:45

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