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

  • 2
    $\begingroup$ 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. $\endgroup$
    – AChem
    Jun 1, 2022 at 16:51
  • 1
    $\begingroup$ 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. $\endgroup$
    – Maurice
    Jun 1, 2022 at 17:04
  • 1
    $\begingroup$ 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) . $\endgroup$
    – Poutnik
    Jun 6, 2022 at 7:27
  • 1
    $\begingroup$ But as NaCl solubility does not change much, I guess positive T effect on ion mobility would dominate. $\endgroup$
    – Poutnik
    Jun 6, 2022 at 7:51

1 Answer 1


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 : $Ba^{2+}$ > $Na^+$.

  • 1
    $\begingroup$ Please don't use MathJax for emphasis. Note that chemical element symbols are upright. Taglines such as "Hope it helps" are unnecessary noise. There are several IUPAC standard symbols for different conductivities, but none of them include $k$ (you can look them up in IUPAC Green Book). It would probably make sense to back up your answer by an equation or some literature data. $\endgroup$
    – andselisk
    Jun 6, 2022 at 4:18
  • $\begingroup$ @andselisk Ah thanks again ! Actually I have trouble typing super scripts so use mathjax, also could you please tell me how I can italicise text ? $\endgroup$
    – PSR_123
    Jun 6, 2022 at 5:08
  • $\begingroup$ @andselisk also, “ The thermal conductivity of a material is a measure of its ability to conduct heat. It is commonly denoted by k, k, λ “, from Wikipedia. k(kappa) is used for conductivity. I don’t have any equation or data as such for the question, only reasoning from various sites I extracted. $\endgroup$
    – PSR_123
    Jun 6, 2022 at 5:10
  • $\begingroup$ One can use $\kappa$ to get $\kappa$. // Useful links for text and formula formatting: Notation basics / Formatting of math/chem expressions / upright vs italic // Use plain texts in CH SE titles. $\endgroup$
    – Poutnik
    Jun 6, 2022 at 6:58
  • $\begingroup$ You are welcome. // For more, see Math SE MathJax tutorial. $\endgroup$
    – Poutnik
    Jun 6, 2022 at 7:20

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