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Standard XII

Chemistry

Bleaching Powder

The conductiv...

Question

The conductivities at infinite dilution of $N H_{4} Cl,NaOH$ and $NaCl$ are 130, 218, 120 $oh m^{- 1} c m^{2} e q^{- 1}$ . If equivalent conductance of N/100 solution of $N H_{4} OH$ is 10, then degree of dissociation of $N H_{4} OH$ at this dilution is:

0.005

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0.043

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0.01

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Solution

The correct option is A 0.043
$N H_{4} C l + N a O H ⟶ N H_{4} O H + N a C l$

$∴ \land_{m}^{\infty}$ $_{N H_{4} O H} = \land_{m}^{\infty}$

$_{N H_{4} C l} + \land_{m}^{\infty}$ $N a O H - \land_{m}^{\infty}$ $_{N a C l}$

$\land_{m}^{\infty} = 130 + 218 - 120$
$⟹ \land_{m}^{\infty} = 228$ $s c m^{2} e q^{2}$

$α = \frac{\land_{m}}{\land_{m^{\infty}}} = \frac{10}{228} = 0.0438$

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Similar questions

Q. The conductivity at infinite dilution of

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and

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10

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Q. The conductivities at infinite dilution of

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Equivalent conductance at infinite dilution, $λ^{0}$ of NH $_{4}$ Cl, NaOH and NaCl are 128.0, 217.8 and 109.9 $o h m^{- 1} c m^{2} e q^{- 1}$ respectively. The equivalent conductance of 0.01 N $N H_{4} O H$ is 9.30 $o h m^{- 1} c m^{2} e q^{- 1}$ , then the degree of ionization of $N H_{4} O H$ at this temperature would be:

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The conductivities at infinite dilution of NH4Cl,NaOH and NaCl are 130, 218, 120 ohm - 1cm2eq - 1. If equivalent conductance of N/100 solution of NH4OH is 10, then degree of dissociation of NH4OH at this dilution is:

The correct option is A 0.043NH4Cl+NaOH⟶NH4OH+NaCl∴∧∞m NH4OH=∧∞m NH4Cl+∧∞m NaOH−∧∞m NaCl∧∞m=130+218−120⟹∧∞m=228 scm2eq2α=∧m∧m∞=10228=0.0438

The conductivities at infinite dilution of $N H_{4} Cl,NaOH$ and $NaCl$ are 130, 218, 120 $oh m^{- 1} c m^{2} e q^{- 1}$ . If equivalent conductance of N/100 solution of $N H_{4} OH$ is 10, then degree of dissociation of $N H_{4} OH$ at this dilution is: