QBManager

CHXII03:ELECTROCHEMISTRY

330322 At $25^{\circ} C$ molar conductance of 0.1 molar aqueous solution of ammonium hydroxide is $9 .54 o h m^{- 1} c m^{2} m o l^{- 1}$ and at infinite dilution its molar conductance is $238 o h m^{- 1} c m^{2} m o l^{- 1}$ . The degree of ionisation at the given temperature is

1

40 .800 %

2

2 .080 %

3

20 .800 %

4

4 .008 %

Explanation:

$Λ_{\infty}$ for acetic acid at infinite dilution
$= 238 o h m^{- 1} c m^{2} m o l^{- 1}$
$Λ_{c}$ for acetic acid at
conc.c $= 9 .54 o h m^{- 1} c m^{2} m o l^{- 1}$
$∴ α = \frac{Λ_{c}}{Λ_{\infty}} = \frac{9.54}{238} = 0.00 4008$

CHXII03:ELECTROCHEMISTRY

330325 At $25^{\circ} C$ , the molar conductance at infinite dilution for the strong electrolytes $NaOH, NaCl$ and ${BaCl}_{2}$ are $248 \times 10^{- 4}, 126 \times 10^{- 4}$ and $280 \times 10^{- 4} {Sm}^{2} {mol}^{- 1}$ respectively. $λ_{m}^{\circ} Ba (OH)_{2}$ in ${Sm}^{2} {mol}^{- 1}$ is

1

362 \times 10^{- 4}

2

402 \times 10^{- 4}

3

524 \times 10^{- 4}

4

568 \times 10^{- 4}

Explanation:

Given,
$\begin{aligned} Λ_{m}^{0} (NaOH) = 248 \times 10^{- 4} {Sm}^{2} {mol}^{- 1} \\ Λ_{m}^{0} (NaCl) = 126 \times 10^{- 4} {Sm}^{2} {mol}^{- 1} \\ Λ_{m}^{0} ({BaCl}_{2}) = 280 \times 10^{- 4} {Sm}^{2} {mol}^{- 1} \end{aligned}$
The reaction for the formation of $Ba (OH)_{2}$ can be written as,
${BaCl}_{2} + 2 NaOH \to Ba (OH)_{2} + 2 NaCl$
$∵ Λ_{mBa {(OH)}_{2}}^{0} = Λ_{mBaC l_{2}}^{0} + 2 Λ_{mNaOH}^{0} - 2 Λ_{mNaCl}^{0}$
$∴ Λ_{mBa {(OH)}_{2}}^{0} = (280 \times 10^{- 4}) + (2 \times 248 \times 10^{- 4}) -$
$(2 \times 126 \times 10^{- 4})$
$= (280 + 496 - 252) \times 10^{- 4}$
$= 524 \times 10^{- 4} S m^{2} mo l^{- 1}$

AIIMS - 2018

CHXII03:ELECTROCHEMISTRY

330326 A weak monobasic acid is $5 %$ dissociated in 0.01 M solution limiting molar conductivity of acid at infinite dilution is $4 \times 1 0^{- 2} o h m^{- 1} m^{2} m o l^{- 1}$ . What will be the conductivity of 0.05 M solution of the acid ?

1

8 .94 \times 1 0^{- 6} o h m^{- 1} m^{2} m o l^{- 1}

2

8 .92 \times 1 0^{- 4} o h m^{- 1} m^{2} m o l^{- 1}

3

4 .46 \times 1 0^{- 6} o h m^{- 1} m^{2} m o l^{- 1}

4

2 .23 \times 1 0^{- 5} o h m^{- 1} m^{2} m o l^{- 1}

Explanation:

$K_{a} = C α^{2} = 0 .01 \times {(0 .05)}^{2}$
$= 2 .5 \times 1 0^{- 5}$
$F o r 0 .05 M$
$α = \sqrt{\frac{K_{a}}{C}} = \sqrt{\frac{2 .5 \times 1 0^{- 5}}{0 .05}} = 0 .223$
$Λ_{m} {= α Λ}_{m}^{\infty} = 0 .0223 \times 4 \times 1 0^{- 2}$
$= 8 .92 \times 1 0^{- 4} o h m^{- 1} m^{2} m o l^{- 1}$

CHXII03:ELECTROCHEMISTRY

330327 Molar conductivity of $N H_{4} O H$ can be calculated by the equation,

1

Λ_{N H_{4} O H}^{^{\circ}} {= Λ}_{B a {(O H)}_{2}}^{^{\circ}} {+ Λ}_{N H_{4} C l}^{^{\circ}} {- Λ}_{B a C l_{2}}^{^{\circ}}

2

Λ_{N H_{4} O H}^{^{\circ}} = \frac{Λ_{B a {(O H)}_{2}}^{^{\circ}} {+ 2 Λ}_{N H_{4} C l}^{^{\circ}} {- Λ}_{B a C l_{2}}^{^{\circ}}}{2}

3

Λ_{N H_{4} O H}^{^{\circ}} {= Λ}_{B a C l_{2}}^{^{\circ}} {+ 2 Λ}_{N H_{4} C l}^{^{\circ}} {- Λ}_{B a {(O H)}_{2}}^{^{\circ}}

4

Λ_{N H_{4} O H}^{^{\circ}} = \frac{{2 Λ}_{N H_{4} C l}^{^{\circ}} {+ Λ}_{B a {(O H)}_{2}}^{^{\circ}}}{2}

Explanation:

$Λ_{B a {(O H)}_{2}}^{^{\circ}} {= λ}_{B a^{+ 2}}^{^{\circ}} {+ 2 λ}_{O H^{- 1}}^{^{\circ}} \to (1)$
$Λ_{N H_{4} C l}^{^{\circ}} {= λ}_{N H_{4^{+}}}^{^{\circ}} {+ 2 λ}_{C l^{- 1}}^{^{\circ}} \to (2)$
$Λ_{B a C l_{2}}^{^{\circ}} {= λ}_{B a^{+ 2}}^{^{\circ}} {+ 2 λ}_{C l^{-}}^{^{\circ}} \to (3)$
According to Kohlrausch’s law, from (1), (2), (3)
${2 Λ}_{N H_{4} O H}^{^{\circ}} {= Λ}_{B a {(O H)}_{2}}^{^{\circ}} {+ 2 Λ}_{N H_{4} C l}^{^{\circ}} {- Λ}_{B a C l_{2}}^{^{\circ}}$
$Λ_{N H_{4} O H}^{^{\circ}} = \frac{Λ_{B a {(O H)}_{2}}^{^{\circ}} {+ 2 Λ}_{N H_{4} C l}^{^{\circ}} {- Λ}_{B a C l_{2}}^{^{\circ}}}{2}$

CHXII03:ELECTROCHEMISTRY

330322 At $25^{\circ} C$ molar conductance of 0.1 molar aqueous solution of ammonium hydroxide is $9 .54 o h m^{- 1} c m^{2} m o l^{- 1}$ and at infinite dilution its molar conductance is $238 o h m^{- 1} c m^{2} m o l^{- 1}$ . The degree of ionisation at the given temperature is

1

40 .800 %

2

2 .080 %

3

20 .800 %

4

4 .008 %

Explanation:

$Λ_{\infty}$ for acetic acid at infinite dilution
$= 238 o h m^{- 1} c m^{2} m o l^{- 1}$
$Λ_{c}$ for acetic acid at
conc.c $= 9 .54 o h m^{- 1} c m^{2} m o l^{- 1}$
$∴ α = \frac{Λ_{c}}{Λ_{\infty}} = \frac{9.54}{238} = 0.00 4008$

CHXII03:ELECTROCHEMISTRY

330325 At $25^{\circ} C$ , the molar conductance at infinite dilution for the strong electrolytes $NaOH, NaCl$ and ${BaCl}_{2}$ are $248 \times 10^{- 4}, 126 \times 10^{- 4}$ and $280 \times 10^{- 4} {Sm}^{2} {mol}^{- 1}$ respectively. $λ_{m}^{\circ} Ba (OH)_{2}$ in ${Sm}^{2} {mol}^{- 1}$ is

1

362 \times 10^{- 4}

2

402 \times 10^{- 4}

3

524 \times 10^{- 4}

4

568 \times 10^{- 4}

Explanation:

Given,
$\begin{aligned} Λ_{m}^{0} (NaOH) = 248 \times 10^{- 4} {Sm}^{2} {mol}^{- 1} \\ Λ_{m}^{0} (NaCl) = 126 \times 10^{- 4} {Sm}^{2} {mol}^{- 1} \\ Λ_{m}^{0} ({BaCl}_{2}) = 280 \times 10^{- 4} {Sm}^{2} {mol}^{- 1} \end{aligned}$
The reaction for the formation of $Ba (OH)_{2}$ can be written as,
${BaCl}_{2} + 2 NaOH \to Ba (OH)_{2} + 2 NaCl$
$∵ Λ_{mBa {(OH)}_{2}}^{0} = Λ_{mBaC l_{2}}^{0} + 2 Λ_{mNaOH}^{0} - 2 Λ_{mNaCl}^{0}$
$∴ Λ_{mBa {(OH)}_{2}}^{0} = (280 \times 10^{- 4}) + (2 \times 248 \times 10^{- 4}) -$
$(2 \times 126 \times 10^{- 4})$
$= (280 + 496 - 252) \times 10^{- 4}$
$= 524 \times 10^{- 4} S m^{2} mo l^{- 1}$

AIIMS - 2018

CHXII03:ELECTROCHEMISTRY

330326 A weak monobasic acid is $5 %$ dissociated in 0.01 M solution limiting molar conductivity of acid at infinite dilution is $4 \times 1 0^{- 2} o h m^{- 1} m^{2} m o l^{- 1}$ . What will be the conductivity of 0.05 M solution of the acid ?

1

8 .94 \times 1 0^{- 6} o h m^{- 1} m^{2} m o l^{- 1}

2

8 .92 \times 1 0^{- 4} o h m^{- 1} m^{2} m o l^{- 1}

3

4 .46 \times 1 0^{- 6} o h m^{- 1} m^{2} m o l^{- 1}

4

2 .23 \times 1 0^{- 5} o h m^{- 1} m^{2} m o l^{- 1}

Explanation:

$K_{a} = C α^{2} = 0 .01 \times {(0 .05)}^{2}$
$= 2 .5 \times 1 0^{- 5}$
$F o r 0 .05 M$
$α = \sqrt{\frac{K_{a}}{C}} = \sqrt{\frac{2 .5 \times 1 0^{- 5}}{0 .05}} = 0 .223$
$Λ_{m} {= α Λ}_{m}^{\infty} = 0 .0223 \times 4 \times 1 0^{- 2}$
$= 8 .92 \times 1 0^{- 4} o h m^{- 1} m^{2} m o l^{- 1}$

CHXII03:ELECTROCHEMISTRY

330327 Molar conductivity of $N H_{4} O H$ can be calculated by the equation,

1

Λ_{N H_{4} O H}^{^{\circ}} {= Λ}_{B a {(O H)}_{2}}^{^{\circ}} {+ Λ}_{N H_{4} C l}^{^{\circ}} {- Λ}_{B a C l_{2}}^{^{\circ}}

2

Λ_{N H_{4} O H}^{^{\circ}} = \frac{Λ_{B a {(O H)}_{2}}^{^{\circ}} {+ 2 Λ}_{N H_{4} C l}^{^{\circ}} {- Λ}_{B a C l_{2}}^{^{\circ}}}{2}

3

Λ_{N H_{4} O H}^{^{\circ}} {= Λ}_{B a C l_{2}}^{^{\circ}} {+ 2 Λ}_{N H_{4} C l}^{^{\circ}} {- Λ}_{B a {(O H)}_{2}}^{^{\circ}}

4

Λ_{N H_{4} O H}^{^{\circ}} = \frac{{2 Λ}_{N H_{4} C l}^{^{\circ}} {+ Λ}_{B a {(O H)}_{2}}^{^{\circ}}}{2}

Explanation:

$Λ_{B a {(O H)}_{2}}^{^{\circ}} {= λ}_{B a^{+ 2}}^{^{\circ}} {+ 2 λ}_{O H^{- 1}}^{^{\circ}} \to (1)$
$Λ_{N H_{4} C l}^{^{\circ}} {= λ}_{N H_{4^{+}}}^{^{\circ}} {+ 2 λ}_{C l^{- 1}}^{^{\circ}} \to (2)$
$Λ_{B a C l_{2}}^{^{\circ}} {= λ}_{B a^{+ 2}}^{^{\circ}} {+ 2 λ}_{C l^{-}}^{^{\circ}} \to (3)$
According to Kohlrausch’s law, from (1), (2), (3)
${2 Λ}_{N H_{4} O H}^{^{\circ}} {= Λ}_{B a {(O H)}_{2}}^{^{\circ}} {+ 2 Λ}_{N H_{4} C l}^{^{\circ}} {- Λ}_{B a C l_{2}}^{^{\circ}}$
$Λ_{N H_{4} O H}^{^{\circ}} = \frac{Λ_{B a {(O H)}_{2}}^{^{\circ}} {+ 2 Λ}_{N H_{4} C l}^{^{\circ}} {- Λ}_{B a C l_{2}}^{^{\circ}}}{2}$

CHXII03:ELECTROCHEMISTRY

330322 At $25^{\circ} C$ molar conductance of 0.1 molar aqueous solution of ammonium hydroxide is $9 .54 o h m^{- 1} c m^{2} m o l^{- 1}$ and at infinite dilution its molar conductance is $238 o h m^{- 1} c m^{2} m o l^{- 1}$ . The degree of ionisation at the given temperature is

1

40 .800 %

2

2 .080 %

3

20 .800 %

4

4 .008 %

Explanation:

$Λ_{\infty}$ for acetic acid at infinite dilution
$= 238 o h m^{- 1} c m^{2} m o l^{- 1}$
$Λ_{c}$ for acetic acid at
conc.c $= 9 .54 o h m^{- 1} c m^{2} m o l^{- 1}$
$∴ α = \frac{Λ_{c}}{Λ_{\infty}} = \frac{9.54}{238} = 0.00 4008$

CHXII03:ELECTROCHEMISTRY

330325 At $25^{\circ} C$ , the molar conductance at infinite dilution for the strong electrolytes $NaOH, NaCl$ and ${BaCl}_{2}$ are $248 \times 10^{- 4}, 126 \times 10^{- 4}$ and $280 \times 10^{- 4} {Sm}^{2} {mol}^{- 1}$ respectively. $λ_{m}^{\circ} Ba (OH)_{2}$ in ${Sm}^{2} {mol}^{- 1}$ is

1

362 \times 10^{- 4}

2

402 \times 10^{- 4}

3

524 \times 10^{- 4}

4

568 \times 10^{- 4}

Explanation:

Given,
$\begin{aligned} Λ_{m}^{0} (NaOH) = 248 \times 10^{- 4} {Sm}^{2} {mol}^{- 1} \\ Λ_{m}^{0} (NaCl) = 126 \times 10^{- 4} {Sm}^{2} {mol}^{- 1} \\ Λ_{m}^{0} ({BaCl}_{2}) = 280 \times 10^{- 4} {Sm}^{2} {mol}^{- 1} \end{aligned}$
The reaction for the formation of $Ba (OH)_{2}$ can be written as,
${BaCl}_{2} + 2 NaOH \to Ba (OH)_{2} + 2 NaCl$
$∵ Λ_{mBa {(OH)}_{2}}^{0} = Λ_{mBaC l_{2}}^{0} + 2 Λ_{mNaOH}^{0} - 2 Λ_{mNaCl}^{0}$
$∴ Λ_{mBa {(OH)}_{2}}^{0} = (280 \times 10^{- 4}) + (2 \times 248 \times 10^{- 4}) -$
$(2 \times 126 \times 10^{- 4})$
$= (280 + 496 - 252) \times 10^{- 4}$
$= 524 \times 10^{- 4} S m^{2} mo l^{- 1}$

AIIMS - 2018

CHXII03:ELECTROCHEMISTRY

330326 A weak monobasic acid is $5 %$ dissociated in 0.01 M solution limiting molar conductivity of acid at infinite dilution is $4 \times 1 0^{- 2} o h m^{- 1} m^{2} m o l^{- 1}$ . What will be the conductivity of 0.05 M solution of the acid ?

1

8 .94 \times 1 0^{- 6} o h m^{- 1} m^{2} m o l^{- 1}

2

8 .92 \times 1 0^{- 4} o h m^{- 1} m^{2} m o l^{- 1}

3

4 .46 \times 1 0^{- 6} o h m^{- 1} m^{2} m o l^{- 1}

4

2 .23 \times 1 0^{- 5} o h m^{- 1} m^{2} m o l^{- 1}

Explanation:

$K_{a} = C α^{2} = 0 .01 \times {(0 .05)}^{2}$
$= 2 .5 \times 1 0^{- 5}$
$F o r 0 .05 M$
$α = \sqrt{\frac{K_{a}}{C}} = \sqrt{\frac{2 .5 \times 1 0^{- 5}}{0 .05}} = 0 .223$
$Λ_{m} {= α Λ}_{m}^{\infty} = 0 .0223 \times 4 \times 1 0^{- 2}$
$= 8 .92 \times 1 0^{- 4} o h m^{- 1} m^{2} m o l^{- 1}$

CHXII03:ELECTROCHEMISTRY

330327 Molar conductivity of $N H_{4} O H$ can be calculated by the equation,

1

Λ_{N H_{4} O H}^{^{\circ}} {= Λ}_{B a {(O H)}_{2}}^{^{\circ}} {+ Λ}_{N H_{4} C l}^{^{\circ}} {- Λ}_{B a C l_{2}}^{^{\circ}}

2

Λ_{N H_{4} O H}^{^{\circ}} = \frac{Λ_{B a {(O H)}_{2}}^{^{\circ}} {+ 2 Λ}_{N H_{4} C l}^{^{\circ}} {- Λ}_{B a C l_{2}}^{^{\circ}}}{2}

3

Λ_{N H_{4} O H}^{^{\circ}} {= Λ}_{B a C l_{2}}^{^{\circ}} {+ 2 Λ}_{N H_{4} C l}^{^{\circ}} {- Λ}_{B a {(O H)}_{2}}^{^{\circ}}

4

Λ_{N H_{4} O H}^{^{\circ}} = \frac{{2 Λ}_{N H_{4} C l}^{^{\circ}} {+ Λ}_{B a {(O H)}_{2}}^{^{\circ}}}{2}

Explanation:

$Λ_{B a {(O H)}_{2}}^{^{\circ}} {= λ}_{B a^{+ 2}}^{^{\circ}} {+ 2 λ}_{O H^{- 1}}^{^{\circ}} \to (1)$
$Λ_{N H_{4} C l}^{^{\circ}} {= λ}_{N H_{4^{+}}}^{^{\circ}} {+ 2 λ}_{C l^{- 1}}^{^{\circ}} \to (2)$
$Λ_{B a C l_{2}}^{^{\circ}} {= λ}_{B a^{+ 2}}^{^{\circ}} {+ 2 λ}_{C l^{-}}^{^{\circ}} \to (3)$
According to Kohlrausch’s law, from (1), (2), (3)
${2 Λ}_{N H_{4} O H}^{^{\circ}} {= Λ}_{B a {(O H)}_{2}}^{^{\circ}} {+ 2 Λ}_{N H_{4} C l}^{^{\circ}} {- Λ}_{B a C l_{2}}^{^{\circ}}$
$Λ_{N H_{4} O H}^{^{\circ}} = \frac{Λ_{B a {(O H)}_{2}}^{^{\circ}} {+ 2 Λ}_{N H_{4} C l}^{^{\circ}} {- Λ}_{B a C l_{2}}^{^{\circ}}}{2}$

CHXII03:ELECTROCHEMISTRY

330322 At $25^{\circ} C$ molar conductance of 0.1 molar aqueous solution of ammonium hydroxide is $9 .54 o h m^{- 1} c m^{2} m o l^{- 1}$ and at infinite dilution its molar conductance is $238 o h m^{- 1} c m^{2} m o l^{- 1}$ . The degree of ionisation at the given temperature is

1

40 .800 %

2

2 .080 %

3

20 .800 %

4

4 .008 %

Explanation:

$Λ_{\infty}$ for acetic acid at infinite dilution
$= 238 o h m^{- 1} c m^{2} m o l^{- 1}$
$Λ_{c}$ for acetic acid at
conc.c $= 9 .54 o h m^{- 1} c m^{2} m o l^{- 1}$
$∴ α = \frac{Λ_{c}}{Λ_{\infty}} = \frac{9.54}{238} = 0.00 4008$

CHXII03:ELECTROCHEMISTRY

330325 At $25^{\circ} C$ , the molar conductance at infinite dilution for the strong electrolytes $NaOH, NaCl$ and ${BaCl}_{2}$ are $248 \times 10^{- 4}, 126 \times 10^{- 4}$ and $280 \times 10^{- 4} {Sm}^{2} {mol}^{- 1}$ respectively. $λ_{m}^{\circ} Ba (OH)_{2}$ in ${Sm}^{2} {mol}^{- 1}$ is

1

362 \times 10^{- 4}

2

402 \times 10^{- 4}

3

524 \times 10^{- 4}

4

568 \times 10^{- 4}

Explanation:

Given,
$\begin{aligned} Λ_{m}^{0} (NaOH) = 248 \times 10^{- 4} {Sm}^{2} {mol}^{- 1} \\ Λ_{m}^{0} (NaCl) = 126 \times 10^{- 4} {Sm}^{2} {mol}^{- 1} \\ Λ_{m}^{0} ({BaCl}_{2}) = 280 \times 10^{- 4} {Sm}^{2} {mol}^{- 1} \end{aligned}$
The reaction for the formation of $Ba (OH)_{2}$ can be written as,
${BaCl}_{2} + 2 NaOH \to Ba (OH)_{2} + 2 NaCl$
$∵ Λ_{mBa {(OH)}_{2}}^{0} = Λ_{mBaC l_{2}}^{0} + 2 Λ_{mNaOH}^{0} - 2 Λ_{mNaCl}^{0}$
$∴ Λ_{mBa {(OH)}_{2}}^{0} = (280 \times 10^{- 4}) + (2 \times 248 \times 10^{- 4}) -$
$(2 \times 126 \times 10^{- 4})$
$= (280 + 496 - 252) \times 10^{- 4}$
$= 524 \times 10^{- 4} S m^{2} mo l^{- 1}$

AIIMS - 2018

CHXII03:ELECTROCHEMISTRY

330326 A weak monobasic acid is $5 %$ dissociated in 0.01 M solution limiting molar conductivity of acid at infinite dilution is $4 \times 1 0^{- 2} o h m^{- 1} m^{2} m o l^{- 1}$ . What will be the conductivity of 0.05 M solution of the acid ?

1

8 .94 \times 1 0^{- 6} o h m^{- 1} m^{2} m o l^{- 1}

2

8 .92 \times 1 0^{- 4} o h m^{- 1} m^{2} m o l^{- 1}

3

4 .46 \times 1 0^{- 6} o h m^{- 1} m^{2} m o l^{- 1}

4

2 .23 \times 1 0^{- 5} o h m^{- 1} m^{2} m o l^{- 1}

Explanation:

$K_{a} = C α^{2} = 0 .01 \times {(0 .05)}^{2}$
$= 2 .5 \times 1 0^{- 5}$
$F o r 0 .05 M$
$α = \sqrt{\frac{K_{a}}{C}} = \sqrt{\frac{2 .5 \times 1 0^{- 5}}{0 .05}} = 0 .223$
$Λ_{m} {= α Λ}_{m}^{\infty} = 0 .0223 \times 4 \times 1 0^{- 2}$
$= 8 .92 \times 1 0^{- 4} o h m^{- 1} m^{2} m o l^{- 1}$

CHXII03:ELECTROCHEMISTRY

330327 Molar conductivity of $N H_{4} O H$ can be calculated by the equation,

1

Λ_{N H_{4} O H}^{^{\circ}} {= Λ}_{B a {(O H)}_{2}}^{^{\circ}} {+ Λ}_{N H_{4} C l}^{^{\circ}} {- Λ}_{B a C l_{2}}^{^{\circ}}

2

Λ_{N H_{4} O H}^{^{\circ}} = \frac{Λ_{B a {(O H)}_{2}}^{^{\circ}} {+ 2 Λ}_{N H_{4} C l}^{^{\circ}} {- Λ}_{B a C l_{2}}^{^{\circ}}}{2}

3

Λ_{N H_{4} O H}^{^{\circ}} {= Λ}_{B a C l_{2}}^{^{\circ}} {+ 2 Λ}_{N H_{4} C l}^{^{\circ}} {- Λ}_{B a {(O H)}_{2}}^{^{\circ}}

4

Λ_{N H_{4} O H}^{^{\circ}} = \frac{{2 Λ}_{N H_{4} C l}^{^{\circ}} {+ Λ}_{B a {(O H)}_{2}}^{^{\circ}}}{2}

Explanation:

$Λ_{B a {(O H)}_{2}}^{^{\circ}} {= λ}_{B a^{+ 2}}^{^{\circ}} {+ 2 λ}_{O H^{- 1}}^{^{\circ}} \to (1)$
$Λ_{N H_{4} C l}^{^{\circ}} {= λ}_{N H_{4^{+}}}^{^{\circ}} {+ 2 λ}_{C l^{- 1}}^{^{\circ}} \to (2)$
$Λ_{B a C l_{2}}^{^{\circ}} {= λ}_{B a^{+ 2}}^{^{\circ}} {+ 2 λ}_{C l^{-}}^{^{\circ}} \to (3)$
According to Kohlrausch’s law, from (1), (2), (3)
${2 Λ}_{N H_{4} O H}^{^{\circ}} {= Λ}_{B a {(O H)}_{2}}^{^{\circ}} {+ 2 Λ}_{N H_{4} C l}^{^{\circ}} {- Λ}_{B a C l_{2}}^{^{\circ}}$
$Λ_{N H_{4} O H}^{^{\circ}} = \frac{Λ_{B a {(O H)}_{2}}^{^{\circ}} {+ 2 Λ}_{N H_{4} C l}^{^{\circ}} {- Λ}_{B a C l_{2}}^{^{\circ}}}{2}$

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