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Electrochemistry question

2023 · 1 Feb · Shift 2 · Q17
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Electrochemistry question

2023 · 1 Feb · Shift 2 · Q17

JEE MainChemistryElectrochemistryNumerical+4 / −1
1×10−5 M AgNO31 \times 10^{-5} ~\mathrm{M} ~\mathrm{AgNO}_{3}1×10−5 M AgNO3​ is added to 1 L1 \mathrm{~L}1 L of saturated solution of AgBr\mathrm{AgBr}AgBr. The conductivity of this solution at 298 K298 \mathrm{~K}298 K is ‾\underline{\hspace{2cm}}​×10−8 S m−1\times 10^{-8} \mathrm{~S} \mathrm{~m}^{-1}×10−8 S m−1. [Given : KSP(AgBr)=4.9×10−13\mathrm{K}_{\mathrm{SP}}(\mathrm{AgBr})=4.9 \times 10^{-13}KSP​(AgBr)=4.9×10−13 at 298 K298 \mathrm{~K}298 KλAg+0=6×10−3 S m2 mol−1λBr−0=8×10−3 S m2 mol−1λNO3−0=7×10−3 S m2 mol−1]\begin{aligned} & \lambda_{\mathrm{Ag}^{+}}^{0}=6 \times 10^{-3} \mathrm{~S} \mathrm{~m}^{2} \mathrm{~mol}^{-1} \\ & \lambda_{\mathrm{Br}^{-}}^{0}=8 \times 10^{-3} \mathrm{~S} \mathrm{~m}^{2} \mathrm{~mol}^{-1} \\ & \left.\lambda_{\mathrm{NO}_{3}^{-}}^{0}=7 \times 10^{-3} \mathrm{~S} \mathrm{~m}^{2} \mathrm{~mol}^{-1}\right] \end{aligned}​λAg+0​=6×10−3 S m2 mol−1λBr−0​=8×10−3 S m2 mol−1λNO3−​0​=7×10−3 S m2 mol−1]​
Numerical answer
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