Question

8.
From the following molar conductivities at infinite dilution,
Am for Al2(SO4)3 = 858 S cm2 mol-'
An for NH,OH = 238.3 S cm mol-1
An for (NH),SO, = 238.4 S cm2 mol-1
Calculate Am for Al(OH)2
​

Answer 1

according to Kohlrausch's law, molar conductivity at infinite dilution can be broken into its ions.

so, $\Lambda_m$ for Al(OH)3 = $\Lambda_m$ for Al^+ + 2 × $\Lambda_m$ for (OH)^-

= 1/2 × $\Lambda_m$ for Al2(SO4)3 + 3 × $\Lambda_m$ for NH4(OH) - 3/2 × $\Lambda_m$ for (NH4)2SO4

given,

$\Lambda_m$ for Al2(SO4)3 = 858 S.cm²/mol

$\Lambda_m$ for NH4OH = 238.3 S cm²/mol

$\Lambda_m$ for (NH4)2SO4 = 238.4 S cm²/mol

now, $\Lambda_m$ for Al(OH)3 = 1/2 × 858 + 3 × 238.3 - 3/2 × 238.4

= 786.3 S cm²/mol

hence, option (3) is correct choice

Answer 2

Answer:

Molar conductivities at infinite dilution :

Λ

m

0

for Al

2

(SO

4

)

3

=858 Scm

2

mol

−1

Λ

m

0

for NH

4

OH=238.3 Scm

2

mol

−1

Λ

m

0

for (NH

4

)

2

SO

4

=238.4 Scm

2

mol

−1

Al

2

(SO

4

)

3

⇌2Al

3+

+3SO

4

2−

NH

4

OH⇌NH

4

+

+OH

−

(NH

4

)

2

SO

4

⇌2NH

4

+

+SO

4

2−

Al(OH)

3

⇌Al

3+

+3OH

−

2(Λ

m

0

Al(OH)

3

)=Λ

m

0

Al

2

(SO

4

)

3

+6(Λ

m

0

NH

4

OH)−3(Λ

m

0

(NH

4

)

2

SO

4

)

2(Λ

m

0

Al(OH)

3

)=858+6(238.3)−3(238.4)

=1572.6

Λ

m

0

Al(OH)

3

=

2

1572.6

=786.3 Scm

2

mol

−1

Hence, option C is correct.