
A \[0.15\] molal solution of \[{{K}_{4}}\left[ Fe{{\left( CN \right)}_{6}} \right]\] in water freezes at \[-0.65{}^\circ C\]. What is the apparent percentage of dissociation of this compound in this solution? [\[{{K}_{f}}\] for water \[=1.86^oC{ }mo{{l}^{-1}}\]]
A.\[0.33\]
B.\[0.52\]
C.\[0.63\]
D.\[0.79\]
Answer
575.1k+ views
Hint:The freezing point of a substance is the temperature at which the liquid is converted to solid form.
-At first we write the balanced chemical equation which takes place in the process. Then we find out the total number of moles in the equilibrium. After that we substitute the values of every variable given in the question and find out the value of van’t hoff’s factor. The percentage of dissociation can be calculated by using van’t hoff’s factor, as it is the extent of dissociation or association which takes place in a reaction.
Formula used: \[\Delta {{T}_{f}}=i{{K}_{f}}m\]
\[\Delta {{T}_{f}}\] denotes the change in freezing point of the solution. \[i\] is the van’t hoff's factor. \[{{K}_{f}}\] molal freezing-point depression constant, and \[m\] stands for the concentration of the solution.
Complete step by step answer:
Freezing point can be defined as the temperature at which a liquid changes to become a solid. Just like the melting point, if we increase the pressure it will usually raise the freezing point. Generally in the case of mixtures and for certain organic compounds such as fats the value of freezing point is lower than that of the melting point. As the mixture of compounds freezes, the solid that forms first usually differs in composition from that of the considered liquid, and formation of the solid results in change in the composition of the liquid left in the mixture, generally in a way that gradually lowers the freezing point. This principle has a usage in purifying mixtures, successive freezing and melting for gradually separating the components. The heat of fusion or the thermal fusion, meaning the heat that must be applied in order to melt a certain solid, must be removed from the liquid in order to freeze it. Some liquids can be converted to supercooled liquid i.e., they are cooled below the temperature of freezing point without formation of solid crystals.
The reaction involved in the following question is,
\[{{K}_{4}}\left[ Fe{{\left( CN \right)}_{6}} \right]\leftrightharpoons 4{{K}^{+}}+{{\left[ Fe{{\left( CN \right)}_{6}} \right]}^{4-}}\]
If you consider the given question, the value of freezing point is given, along with the concentration of the solution and the freezing constant. We are supposed to find the percentage of dissociation constant, for that we will find out the dissociation constant by using the equation,
\[\Delta {{T}_{f}}=i{{K}_{f}}m\]
Where \[~\Delta {{T}_{f}}\] is the freezing point which is given in the question, \[-0.65{}^\circ C\]
\[{{K}_{f}}\] Is molal freezing-point depression constant whose value is also known from the given question, \[1.86\circ C{ }mo{{l}^{-1}}\]
And \[m\] stands for the concentration of the solution, which is known to us, \[0.15\]
And \[~i\] is the van’t hoff’s factor, with which we can calculate the value of the dissociation constant which we have to find.
\[{{K}_{4}}\left[ Fe{{\left( CN \right)}_{6}} \right]\to 4{{K}^{\oplus }}+[Fe{{\left( CN{{)}_{6}} \right]}^{4-}}\]
Total moles at equilibrium \[=1-\alpha +4\alpha +\alpha =1+4\alpha \]
So now substituting the values in the following equation, we get
\[0.65=i\times 1.86\times 0.15\]
We know that van’t hoff’s factor is used to express the extent of association or dissociation, so now we solve this equation, for the value of \[~i\], we get
\[i=2.33=1+4\alpha \]
Where, \[\alpha \] is the dissociation constant.
So the value of \[\alpha \] comes out to be, \[0.33\].
So the correct answer is option A.
Note:
-Depression in freezing point is a phenomena that generally occurs when a non-volatile solute is added to the solvent. For instance, salt as a solute in water as a solvent.
-The van’t Hoff’s factor is the ratio of the actual concentration of the particles which are produced when the substance is dissolved and the concentration of a substance as calculated from its mass.
-At first we write the balanced chemical equation which takes place in the process. Then we find out the total number of moles in the equilibrium. After that we substitute the values of every variable given in the question and find out the value of van’t hoff’s factor. The percentage of dissociation can be calculated by using van’t hoff’s factor, as it is the extent of dissociation or association which takes place in a reaction.
Formula used: \[\Delta {{T}_{f}}=i{{K}_{f}}m\]
\[\Delta {{T}_{f}}\] denotes the change in freezing point of the solution. \[i\] is the van’t hoff's factor. \[{{K}_{f}}\] molal freezing-point depression constant, and \[m\] stands for the concentration of the solution.
Complete step by step answer:
Freezing point can be defined as the temperature at which a liquid changes to become a solid. Just like the melting point, if we increase the pressure it will usually raise the freezing point. Generally in the case of mixtures and for certain organic compounds such as fats the value of freezing point is lower than that of the melting point. As the mixture of compounds freezes, the solid that forms first usually differs in composition from that of the considered liquid, and formation of the solid results in change in the composition of the liquid left in the mixture, generally in a way that gradually lowers the freezing point. This principle has a usage in purifying mixtures, successive freezing and melting for gradually separating the components. The heat of fusion or the thermal fusion, meaning the heat that must be applied in order to melt a certain solid, must be removed from the liquid in order to freeze it. Some liquids can be converted to supercooled liquid i.e., they are cooled below the temperature of freezing point without formation of solid crystals.
The reaction involved in the following question is,
\[{{K}_{4}}\left[ Fe{{\left( CN \right)}_{6}} \right]\leftrightharpoons 4{{K}^{+}}+{{\left[ Fe{{\left( CN \right)}_{6}} \right]}^{4-}}\]
If you consider the given question, the value of freezing point is given, along with the concentration of the solution and the freezing constant. We are supposed to find the percentage of dissociation constant, for that we will find out the dissociation constant by using the equation,
\[\Delta {{T}_{f}}=i{{K}_{f}}m\]
Where \[~\Delta {{T}_{f}}\] is the freezing point which is given in the question, \[-0.65{}^\circ C\]
\[{{K}_{f}}\] Is molal freezing-point depression constant whose value is also known from the given question, \[1.86\circ C{ }mo{{l}^{-1}}\]
And \[m\] stands for the concentration of the solution, which is known to us, \[0.15\]
And \[~i\] is the van’t hoff’s factor, with which we can calculate the value of the dissociation constant which we have to find.
\[{{K}_{4}}\left[ Fe{{\left( CN \right)}_{6}} \right]\to 4{{K}^{\oplus }}+[Fe{{\left( CN{{)}_{6}} \right]}^{4-}}\]
| Initial | $1$ | $0$ | $0$ |
| At eq. | \[1-\alpha \] | \[4\alpha \] | \[\alpha \] |
Total moles at equilibrium \[=1-\alpha +4\alpha +\alpha =1+4\alpha \]
So now substituting the values in the following equation, we get
\[0.65=i\times 1.86\times 0.15\]
We know that van’t hoff’s factor is used to express the extent of association or dissociation, so now we solve this equation, for the value of \[~i\], we get
\[i=2.33=1+4\alpha \]
Where, \[\alpha \] is the dissociation constant.
So the value of \[\alpha \] comes out to be, \[0.33\].
So the correct answer is option A.
Note:
-Depression in freezing point is a phenomena that generally occurs when a non-volatile solute is added to the solvent. For instance, salt as a solute in water as a solvent.
-The van’t Hoff’s factor is the ratio of the actual concentration of the particles which are produced when the substance is dissolved and the concentration of a substance as calculated from its mass.
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