Respuesta :
Answer:
The degree of dissociation is 1
thus the fraction dissociated is one
Explanation:
The depression in freezing point is a colligative property and depends upon the number of solute particles present in the solution.
The relation between depression in freezing point and molality is:
[tex]depressioninfreezingpoint=(i)K_{f}Xmolality[/tex]
For water
Kf=1.86 °C/m
Where
i= Van't Hoff factor
For electrolytes the i depends upon on the extent of dissociation
i = αn + (1 - α)
Where
α = is degree of dissociation
Let us put the values
[tex]moles=\frac{mass}{molarmass}=\frac{4}{47}=0.085[/tex]
[tex]0.1692=i\frac{0.0851}{1}\\ i=2[/tex]
Putting value:
[tex]2 = \alpha n + (1 - \alpha )\\n=2\\\alpha =1[/tex]
The fraction of HNO₂ that has dissociated is 1.
Freezing point depression:
It is a colligative property observed in solutions that results from the introduction of solute molecules to a solvent. It is given by:
ΔT =kf * i *m
For water:
Kf=1.86 °C/m
For electrolytes the "i" depends upon on the extent of dissociation:
i = αn + (1 - α)
where
α = degree of dissociation
On substituting the values:
[tex]\text{Number of moles}=\frac{\text{Given mas}}{\text{Molar mass}}\\\\\text{Number of moles}=\frac{4g}{47g/mol}\\\\ \text{Number of moles}=0.085 moles[/tex]
→ Calculation of "i":
[tex]0.01692=i*\frac{0.0851}{1}\\\\ i=2[/tex]
→ Calculation of "α:
[tex]2\alpha=\alpha n+(1-\alpha)\\\\n=2\\\\\alpha=1[/tex]
Thus, the value of dissociation constant is 1 and the fraction of HNO₂ that has dissociated is 1.
Find more information about Dissociation constant here:
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