At 300 K, the molar conductivities of the aqueous solutions of three salts at two different concentrations are given below: Salt Concentration (M) Molar conductivity (S cm^2 mol^ -1 ) NaNO _3 0.01 111 0.04 101 NaCl 0.01 117 0.04 107 AgNO _3 0.01 125 0.04 116 The conductivity of a saturated aqueous s

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At 300 K, the molar conductivities of the aqueous solutions of three salts at two different concentrations are given below: Salt Concentration (M) Molar conductivity (S cm^2 mol^ -1 ) NaNO _3 0.01 111 0.04 101 NaCl 0.01 117 0.04 107 AgNO _3 0.01 125 0.04 116 The conductivity of a saturated aqueous solution of AgCl is 1.40 10^ -6 S cm^ -1 at 300 K. If the solubility of AgCl in water at 300 K is X mol L^ -1 , then _ 10 (X^ -1 ) is (Assume that AgCl dissolved in water ionizes completely and that the molar conductivity of saturated AgCl solution is equal to its limiting molar conductivity.)

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Correct answer: C. 5

Using the Debye-Hückel-Onsager equation, _m = ^ _m - b C , we can find the limiting molar conductivity ( ^ _m) for each salt. For NaNO _3: ^ _m - b 0.01 = 111 ^ _m - 0.1b = 111 ^ _m - b 0.04 = 101 ^ _m - 0.2b = 101 Subtracting the two equations gives 0.1b = 10 b = 100. ^ _m( NaNO _3) = 111 + 100(0.1) = 121 S cm ^2 mol ^ -1 . For NaCl : ^ _m - 0.1b = 117 ^ _m - 0.2b = 107 Subtracting gives 0.1b = 10 b = 100. ^ _m( NaCl ) = 117 + 100(0.1) = 127 S cm ^2 mol ^ -1 . For AgNO _3: ^ _m - 0.1b = 125 ^ _m - 0.2b = 116 Subtracting gives 0.1b = 9 b = 90. ^ _m( AgNO _3) = 125 + 90(0.1) = 134 S cm ^2 mol ^ -1 . According to Kohlrausch's law of independent migration of ions: ^ _m( AgCl ) = ^ _m( AgNO _3) + ^ _m( NaCl ) - ^ _m( NaNO _3) ^ _m( AgCl ) = 134 + 127 - 121 = 140 S cm ^2 mol ^ -1 . The molar conductivity of the saturated AgCl solution is equal to its limiting molar conductivity, so _m( AgCl ) = 140 S cm ^2 mol ^ -1 . The relationship between molar conductivity, conductivity ( ), and solubility (X in mol L ^ -1 ) is: _m = 1000 X Substituting the given values: 140 = 1.40 10^ -6 1000 X 140 = 1.40 10^ -3 X X = 1.40 10^ -3 140 = 10^ -5 mol L ^ -1 . We need to find _ 10 (X^ -1 ): X^ -1 = 1 10^ -5 = 10^5 _ 10 (X^ -1 ) = _ 10 (10^5) = 5. Answer: 5

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