Question

At ${{25}^{0}}C$ molar conductance of 0.1 molar aqueous solution of ammonium hydroxide is $9.54oh{{m}^{-1}}c{{m}^{2}}mo{{l}^{-1}}$ and at infinite dilution its molar conductance is $238oh{{m}^{-1}}c{{m}^{2}}mo{{l}^{-1}}$. The degree of ionisation of ammonium hydroxide at the same concentration and temperature is:
A. $2.080%$
B. $20.800%$
C. $4.008%$
D. $40.800%$

Answer 1

Hint: You have been provided the molar conductance at 0.2 molar conductance and molar conductance at infinite dilution. Try to find out the degree of ionisation using these 2 values and multiply the answer by 100 to get the percentage of ionisation.

Complete step by step solution:
Let us first know some facts about molar conductivity. Molar conductivity is the conductance property of an answer which contains one mole of the electrolyte or it's a function of the ionic strength of the concentration of salt.
In other words, molar conductivity also can be defined because the conducting power of all the ions that are formed by dissolving a mole of electrolyte in a solution. Molar conductivity is the property of an electrolyte solution that's mainly utilized in determining the efficiency of a given electrolyte in conducting electricity during a solution. Limiting molar conductivity is the molar conductivity at infinite dilution or approximately zero concentration of the electrolyte. For a powerful electrolyte the limiting molar conductivity is often determined by extrapolating the molar conductivity v/s concentration curve and it's but unity. While for a weak electrolyte it can't be determined by extrapolating the curve.
The degree of dissociation (also referred to as the degree of ionization), could be a way of representing the strength of an acid. It's defined because of the ratio of the quantity of ionized molecules and also the number of molecules dissolved in water. It will be represented as a decimal number or as a percentage.
 Let us do the calculations:
We have
\[{{\Lambda }^{0}}_{m}=238oh{{m}^{-1}}c{{m}^{2}}mo{{l}^{-1}}\]
\[{{\Lambda }_{m}}=9.54oh{{m}^{-1}}c{{m}^{2}}mo{{l}^{-1}}\], which are the limiting molar conductivity and molar conductivity respectively
Now, degree of dissociation,
\[\alpha =\dfrac{{{\Lambda }_{m}}}{{{\Lambda }^{0}}_{m}}\]
$={\dfrac{9.54}{2.38}}$
$=$ \[4.008\times {{10}^{-2}}\]

So, the percentage dissociation is 4.008%, which gives option C as the answer.

Note: Always check for the units in the question and convert dimensions to SI units in order to avoid calculation errors. Degree of dissociation is a dimensionless quantity.