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Standard XII

Chemistry

Concentration Terms - An Introduction (w/w etc)

At 18∘C, th...

Question

At $18^{\circ} C$ , the mobilities of $N H_{4}^{+}$ and $C l O_{4}$ $^{-}$ ions are $6.6 \times 10^{- 4}$ and $5.7 \times 10^{- 4} c m^{2} v o l t^{- 1} s e c^{- 1}$ at infinite dilution. Calculate the equivalent conductance of ammonium chlorate solution.

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Solution

$\land_{N H_{4} C l O_{4}}^{\infty} = λ_{N H_{4}^{+}}^{\infty} + λ_{C l O_{4}^{-}}^{\infty}$
$= (U_{N H_{4}^{+}} \infty + U_{C l O_{4}^{-}} \infty) F$
$= (6.6 \times 10^{- 4} + 5.7 \times 10^{- 4}) \times 96500$
$= 118.69 o h m^{- 1} c m^{2} e q^{- 1}$ .

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At 18∘C, the mobilities of NH+4 and ClO4− ions are 6.6×10−4 and 5.7×10−4cm2volt−1sec−1 at infinite dilution. Calculate the equivalent conductance of ammonium chlorate solution.

∧∞NH4ClO4=λ∞NH+4+λ∞ClO−4=(UNH+4∞+UClO−4∞)F=(6.6×10−4+5.7×10−4)×96500=118.69 ohm−1cm2eq−1.

At $18^{\circ} C$ , the mobilities of $N H_{4}^{+}$ and $C l O_{4}$ $^{-}$ ions are $6.6 \times 10^{- 4}$ and $5.7 \times 10^{- 4} c m^{2} v o l t^{- 1} s e c^{- 1}$ at infinite dilution. Calculate the equivalent conductance of ammonium chlorate solution.

$\land_{N H_{4} C l O_{4}}^{\infty} = λ_{N H_{4}^{+}}^{\infty} + λ_{C l O_{4}^{-}}^{\infty}$
$= (U_{N H_{4}^{+}} \infty + U_{C l O_{4}^{-}} \infty) F$
$= (6.6 \times 10^{- 4} + 5.7 \times 10^{- 4}) \times 96500$
$= 118.69 o h m^{- 1} c m^{2} e q^{- 1}$ .