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The following question was asked in JEE Mains 2020:

Let $C_{\ce{NaCl}}$ and $C_{\ce{BaSO4}}$ be the conductances (in S) measured for saturated aqueous solutions of $\ce{NaCl}$ and $\ce{BaSO4}$, respectively, at a temperature $T$. Which of the following is false?

(A) Ionic mobilities of ions from both salts increase with $T$
(B) $C_{\ce{NaCl}} >> C_{\ce{BaSO4}}$ at a given $T$
(C) $C_{\ce{NaCl}}$($T_2$) > $C_{\ce{NaCl}}$($T_1$) for $T_2 > T_1$
(D) $C_{\ce{BaSO4}}$($T_2$) > $C_{\ce{BaSO4}}$($T_1$) for $T_2 > T_1$

My attempt:

(A) As temperature increases viscosity decreases leading to higher mobility, hence this is true.

(B) $\ce{BaSO4}$ is almost insoluble in water, hence its saturated solution will have a much smaller quantity of ions compared to an $\ce{NaCl}$ saturated solution, hence this is true.

(C) As can be seen from the graph here, the solublity of $\ce{NaCl}$ does increase with temperature, but only slightly; nevertheless this option is also true because it does increase and also ionic mobility increases.

(D) Dissolution of $\ce{BaSO4}$ is an endothermic process, hence I would expect solubility (hence conductance) to increase with temperature (although I was not able to find any data), therefore this option is also true.

Where am I going wrong? This is a single-correct question. The given answer is:

(A), but there are varied options found all over the web for this question

Box Box Box Box
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    The answers are self-contradictory. If C and D are correct, then A must be correct. It cannot be false. Ion mobilities increase with temperature in solutions. I don't know what the examiner was thinking when writing this question. – AChem Jun 01 '22 at 16:51
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    Barium sulfate solubility increases with temperature. Here values from Kohlrausch : $\pu{3°C - 0.207 mg/L ; 18°C - 0.23 mg/L ; 26°C - 0.266 mg/L ; 34°C - 0.29 mg/L}$ These values are cited in a recent report : Arbaoui M. Ali, Energy Procedia 157 (2019) 879 - 891. – Maurice Jun 01 '22 at 17:04
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    There can be catch if they mean NaCl solution is saturated at all T ( with excess of solid salt) or if just saturated solution is taken and heated up, becoming unsaturated. For the former, there can be competition of mobility increase due T versus effective mobility decrease for big electrolyte concentrations ( due ion atmosphere and ionic pair formation) . – Poutnik Jun 06 '22 at 07:27
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    But as NaCl solubility does not change much, I guess positive T effect on ion mobility would dominate. – Poutnik Jun 06 '22 at 07:51

1 Answers1

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Okay, there is actually a trick way to solve the question. If any of A, C or D are false the one of the others also becomes false. (As mentioned by @AChem).
So the only option left is (B), which should (ideally) be the correct answer.

Now, coming to the actual reason, as per what I’ve found, conductivity ($k$) of salts in $aqueous$ solution depends on their concentration and number of ions generated on dissociation.

$k$ in aqueous solution or water : $Ba^{2+}$ > $Na^+$.
$k$ in molten state : $Ba^{2+}$ > $Na^+$.

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