Question

The specific conductivity of 0.1M NH,OH is 3.6 *10-4 ohm'cm
. The molar ionic conductance at infinite dilution for NH4+
ion and OH ion are 53 and 198 ohm-'cm-mol respectively calculate
A. Molar conductance of the solution
B. Degree of dissociation of 0.1M NH,OH​

Answer 1

Final answer:

A. Molar conductance of the solution = 3.6 $ohm^{-1} cm^{2} mol^{-1}$

B. Degree of dissociation of 0.1 M $NH_{4}OH$ = 0.01434

Given that: We are given the specific conductivity of 0.1 M $NH_{4}OH$ is $3.6 \times 10^{-4} \ ohm^{-1} cm^{-1}$. The molar ionic conductance at infinite dilution for $NH_{4}^{+}$ ion and $OH^{-}$ ion are $53 \ ohm^{-1} \ cm^{2} \ mol^{-1}$ and $198 \ ohm^{-1} \ cm^{2} \ mol^{-1}$ respectively.

To find: We have to find,

A. Molar conductance of the solution.

B. Degree of dissociation of 0.1 M $NH_{4}OH$.

Explanation:

• The specific conductivity nowadays referred to as conductivity, of any conductor is the reciprocal of the resistivity and is represented by the symbol "K".

A. Molar conductance of the solution:

• If M is the molarity (i.e., concentration in $mol \ dm^{-3}$ or $mol \ L^{-1}$) of the solution and the value of K is expressed in $ohm^{-1} cm^{-1}$, then,

Molar conductivity ($\Lambda_m^c}$) = $\frac{K \times 1000}{M}$

Here, The specific conductivity (K) of 0.1 M $NH_{4}OH$

$=3.6 \times 10^{-4} \ ohm^{-1} cm^{-1}$

Molarity (M) = 0.1 M

• Substitute the values in the equation of molar conductivity.

Molar conductivity ($\Lambda_{m}^{c}$) =$\frac{(3.6 \times 10^{-4} \ ohm^{-1}cm^{-1}) \times (1000 \ cm^{3} dm^{-3})}{(0.1 \ mol \ dm^{-3})}$

$\Lambda_{m}^{c}=3.6 \ ohm^{-1} \ cm^{2} \ mol^{-1}$

Hence, the molar conductance of the solution $=3.6 \ ohm^{-1} \ cm^{2} \ mol^{-1}$

B. Degree of dissociation of 0.1 M $NH_{4}OH$$:$

• The molar ionic conductance of $NH_{4}^{+}$ ion at infinite dilution,

$\lambda _{m}^{0}(NH_{4}^{+})=53 \ ohm^{-1}cm^{2}mol^{-1}$   [given]

• The molar ionic conductance of $OH^{-}$ ion at infinite dilution,

$\lambda _{m}^{0}(OH^{-})= 198 \ ohm^{-1} cm^{2} mol^{-1}$  [given]

• The molar conductance at infinite dilution of $NH_{4}OH$,

$\Lambda_{m}^{0}(NH_{4}OH)=$ $\lambda _{m}^{0}(NH_{4}^{+})+ \lambda_{m}^{0}(OH^{-})$

$\Lambda_{m}^{0}(NH_{4}OH)=53 \ ohm^{-1} cm^{2} mol^{-1} + 198 \ ohm^{-1} cm^{2} mol^{-1}$

$\Lambda_{m}^{0}(NH_{4}OH)=251 \ ohm^{-1} \ cm^{2} \ mol^{-1}$

• The molar conductance of 0.1 M $NH_{4}OH$ solution,

$\Lambda_{m}^{c}=3.6 \ ohm^{-1} cm^{2} mol^{-1}$

• Degree of dissociation:

$\alpha=\frac{\Lambda^_{m}^{c}}{ \Lambda_{m}^{0}}$

• Substitute corresponding values.

$\alpha = \frac{3.6 \ ohm^{-1} cm^{2} mol^{-1}}{ 251 \ ohm^{-1}cm^{2} mol^{-1}}\\\\\alpha = 0.01434$

Hence, the degree of dissociation of the given 0.1 M $NH_{4}OH$ solution is 0.01434

To know more about the concept please go through the links

https://brainly.in/question/8115292

https://brainly.in/question/37222305

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