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Question: The molar conductivity at infinite dilution of \(AgN{O_3},NaCl \;and\; NaN{O_3}\) are 116.5, 110.3 a...

The molar conductivity at infinite dilution of AgNO3,NaCl  and  NaNO3AgN{O_3},NaCl \;and\; NaN{O_3} are 116.5, 110.3 and 105.2 S cm2mol−1c{m^2}mo{l^{ - 1}} respectively. The conductivity of AgCl in water is 2.40×10−6Scm−12.40 \times {10^{ - 6}}Sc{m^{ - 1}} and of water used is 1.16×10−6Scm−11.16 \times {10^{ - 6}}Sc{m^{ - 1}} . the solubility of AgCl will be:
A. 1.463×10−3g/L1.463 \times {10^{ - 3}}g/L
B. 1.163×10−2g/L1.163 \times {10^{ - 2}}g/L
C. 1.163×10−3g/L1.163 \times {10^{ - 3}}g/L
D. None of these

Explanation

Solution

Molar conductance is defined as the conductance of the solution due to all the ions produced by one mole of the dissolved electrolyte in a given solution. We can define electrolytic conductance κ\kappa in the similar terms as Λm=1000κM{\Lambda _m} = \dfrac{{1000\kappa }}{M}. Its symbol is Λm{\Lambda _m} and in the SI system it is denoted by Sm2mol−1S{m^2}mo{l^{ - 1}}.

Complete step by step answer:
Upon dilution i.e. lowering concentration, specific conductance decreases while equivalent and molar conductances increase. At infinite or almost zero concentration equivalent conductance and molar conductance attain their respective limiting values called equivalent conductance at infinite dilution or zero concentration and molar conductance at infinite dilution Λm∞\Lambda _m^\infty or zero concentration Λm∘\Lambda _m^ \circ respectively.
From kohlrausch's law of independent migration of ions we know that at infinite dilution when dissociation is complete, every ion makes some definite contribution towards molar conductance of the electrolyte irrespective of the nature of the other ion with which it is associated and that of molar conductance at infinite dilution for any electrolyte is given by the sum of the contribution of the two ions. Thus
Λm∘=λ+∘+λ−∘\Lambda _m^ \circ = \lambda _ + ^ \circ + \lambda _ - ^ \circ
Let us now apply the above concepts to the question,
Λ∞AgNO3=116.5=λAg+∞+λNO3−∞{\Lambda ^\infty }_{AgN{O_3}} = 116.5 = \lambda _{A{g^ + }}^\infty + \lambda _{NO_3^ - }^\infty.............................(I)
Λ∞NaCl=110.3=λNa+∞+λCl−∞{\Lambda ^\infty }_{NaCl} = 110.3 = \lambda _{N{a^ + }}^\infty + \lambda _{C{l^ - }}^\infty ............................(II)
Λ∞NaNO3=105.2=λNa+∞+λNO3−∞{\Lambda ^\infty }_{NaN{O_3}} = 105.2 = \lambda _{N{a^ + }}^\infty + \lambda _{NO_3^ - }^\infty ...................(III)
On Adding the first two equations and subtracting the third equation we will get infinite dilution for silver chloride.
ΛAgCl∞=λAg+∞+λCl−∞=116.5+110.3−105.2=121.6Scm2/eq\Lambda _{AgCl}^\infty = \lambda _{A{g^ + }}^\infty + \lambda _{C{l^ - }}^\infty = 116.5 + 110.3 - 105.2 = 121.6Sc{m^2}/eq
ΛAgCl=κ×1000M{\Lambda _{AgCl}} = \dfrac{{\kappa \times 1000}}{M}
In the question we are given conductivity of AgCl in water and water. So we can calculate conductivity of solution by subtracting conductivity of AgCl in water and water, which will be κAgCl=(2.40−1.16)×10−6=1.24×10−6Scm2 ΛAgCl=1.24×1000×10−6M=121.6 M=1.24×10−3121.6=1.463×10−3g/L {\kappa _{AgCl}} = (2.40 - 1.16) \times {10^{ - 6}} = 1.24 \times {10^{ - 6}}Sc{m^2} \\\ {\Lambda _{AgCl}} = \dfrac{{1.24 \times 1000 \times {{10}^{ - 6}}}}{M} = 121.6 \\\ M = \dfrac{{1.24 \times {{10}^{ - 3}}}}{{121.6}} = 1.463 \times {10^{ - 3}}g/L \\\

Therefore the correct option is A.

Note: As dilution approaches infinity the degree of dissociation of weak electrolyte approaches unity. The decrease in kappa may also be explained as upon dilution the number of ions increases in case of weak electrolyte but its volume also increases.