JEE MAIN - Chemistry (2024 - 5th April Morning Shift - No. 15)
Molar ionic conductivities of divalent cation and anion are $$57 \mathrm{~S~cm}^2 \mathrm{~mol}^{-1}$$ and $$73 \mathrm{~S~cm}^2 \mathrm{~mol}^{-1}$$ respectively. The molar conductivity of solution of an electrolyte with the above cation and anion will be:
$$187 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}$$
$$260 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}$$
$$65 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}$$
$$130 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}$$
Explanation
The compound with divalent cation $$\left(\mathrm{A}^{2+}\right)$$ and anion $$\left(B^{2-}\right)$$ will be $$A B$$.
Molar conductivity of its solution will be
$$57+73=130 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}$$
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