
If relative decrease in vapour pressure is 0.4 for a solution containing 1 mole NaCl in 3 moles of ${H_2}O$, then \[\% \] ionization of NaCl is
A. \[60\% \]
B. \[80\% \]
C. \[40\% \]
D. \[100\% \]
Answer
566.4k+ views
Hint: Since the relative decrease in the vapour pressure is given in the question and we also know the relative decrease in the vapour pressure equals \[\dfrac{{{p^0} - {p^s}}}{{{p^0}}}\]. We will find the value of Van’t hoff factor ‘i’. Now as NaCl will be dissociated into \[N{a^ + }\] and \[C{l^ - }\], so n will be equals 2. Now that we know ‘i’ and ‘n’, we will find the degree of dissociation/ionisation, \[\alpha \].
Complete step by step answer:
Relative decrease in the vapour pressure is given by \[ = \dfrac{{{p^0} - {p^s}}}{{{p^0}}} = 0.4\]
Since, \[\dfrac{{{p^0} - {p^s}}}{{{p^0}}} = \dfrac{{i{n_{solute}}}}{{i{n_{solute}} + i{n_{solvant}}}}\]
Where, i = Van’t hoff factor
Because NaCl will be dissociated into \[N{a^ + }\] and \[C{l^ - }\]
\[NaCl \to N{a^ + } + C{l^ - }\]
\[\dfrac{{{p^0} - {p^s}}}{{{p^0}}} = \dfrac{{i{n_{solute}}}}{{i{n_{solute}} + i{n_{solvant}}}}\]
$ \Rightarrow 0.4 = \dfrac{{i(1)}}{{i(1) + 3i}}$
\[ \Rightarrow 0.4i + 1.2i = i\]
\[ \Rightarrow 1.6i = i\]
\[ \Rightarrow 0.6i = 1\]
\[ \Rightarrow i = 1.6\]
Since, \[n = 2\] (n = number of dissociated or associated particles). In this case it will be dissociated particles.
We know, \[ \alpha = \dfrac{{i - 1}}{{n - 1}}\] (for dissociation)
\[ \alpha = \dfrac{{1 - i}}{{1 - \dfrac{1}{n}}}\] (for association)
\[ \Rightarrow \alpha = \dfrac{{1.6 - 1}}{{1.1}}\]
\[ \Rightarrow \alpha = 0.6\]
\[ \Rightarrow \alpha \% = 60\% \]
So, the \[\% \] ionization of NaCl is \[60\% \]
Therefore, the correct answer is option (A).
Note: The degree of dissociation is the phenomenon of generating current that carries free ions, which get dissociated from the fraction of solute at a given concentration. It is represented by the symbol. The ratio of the actual concentration of the particles produced (solute + solvent) when the substance is dissolved and the concentration of a substance (solute) as calculated from its mass is the van 't hoff factor.
Complete step by step answer:
Relative decrease in the vapour pressure is given by \[ = \dfrac{{{p^0} - {p^s}}}{{{p^0}}} = 0.4\]
Since, \[\dfrac{{{p^0} - {p^s}}}{{{p^0}}} = \dfrac{{i{n_{solute}}}}{{i{n_{solute}} + i{n_{solvant}}}}\]
Where, i = Van’t hoff factor
Because NaCl will be dissociated into \[N{a^ + }\] and \[C{l^ - }\]
\[NaCl \to N{a^ + } + C{l^ - }\]
\[\dfrac{{{p^0} - {p^s}}}{{{p^0}}} = \dfrac{{i{n_{solute}}}}{{i{n_{solute}} + i{n_{solvant}}}}\]
$ \Rightarrow 0.4 = \dfrac{{i(1)}}{{i(1) + 3i}}$
\[ \Rightarrow 0.4i + 1.2i = i\]
\[ \Rightarrow 1.6i = i\]
\[ \Rightarrow 0.6i = 1\]
\[ \Rightarrow i = 1.6\]
Since, \[n = 2\] (n = number of dissociated or associated particles). In this case it will be dissociated particles.
We know, \[ \alpha = \dfrac{{i - 1}}{{n - 1}}\] (for dissociation)
\[ \alpha = \dfrac{{1 - i}}{{1 - \dfrac{1}{n}}}\] (for association)
\[ \Rightarrow \alpha = \dfrac{{1.6 - 1}}{{1.1}}\]
\[ \Rightarrow \alpha = 0.6\]
\[ \Rightarrow \alpha \% = 60\% \]
So, the \[\% \] ionization of NaCl is \[60\% \]
Therefore, the correct answer is option (A).
Note: The degree of dissociation is the phenomenon of generating current that carries free ions, which get dissociated from the fraction of solute at a given concentration. It is represented by the symbol. The ratio of the actual concentration of the particles produced (solute + solvent) when the substance is dissolved and the concentration of a substance (solute) as calculated from its mass is the van 't hoff factor.
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