
Calculate the freezing point of a solution containing 8.1 g of HBr in 100 g of water, assuming the acid to be $ 90\% $ ionized.
[Given: Molar mass of Br = 80 $ g/mol $ , $ {K_f} $ of water = 1.86 K $ g/mol $ ]
Answer
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Hint: For solving the given question, follow these steps: Calculate the number of moles of solute, then calculate the molality. After that calculate the Van’t Hoff factor and the depression in freezing point. In the end, from the depression in freezing point, calculate the new freezing point.
Complete Step-by-Step Answer
Before we move ahead with the solution of this question, let us first understand some basic concepts:
Molality: Molality can be basically explained as the ratio of the number of moles of the solute to the total mass of the solvent. The unit for the mass of the solvent must be ‘kg’.
Molality = $ \dfrac{{number\,\,of\,\,moles\,\,of\,\,solute}}{{mass\,\,of\,\,solvent}} $ =m
The number of moles of HBr can thus be calculated as:
Number of moles = $ \dfrac{{total\,\,mass}}{{molar\,\,mass}} $ = $ \dfrac{{8.1}}{{80}} = 0.10125 $ moles
Molality = m = $ \dfrac{{0.10125}}{{0.100}} $
In aqueous solution, HBr dissociates as follows:
$ HBr \Leftrightarrow {H^ + } + {\mathbf{B}}{r^ - } $
Hence, even though we only have one compound in hand, upon dissociation we will have two ions, two correct this anomaly, we will use Van’t Hoff Factor. The Van’t Hoff factor for this case can be calculated as:
i = 1 + $ \alpha $ (n-1), where $ \alpha $ is the degree of dissociation and n is the number of ions formed
i = 1 + (2-1) (0.9) = 1.9
Now, the change in freezing point can be calculated as:
$ \Delta {T_f} = $ (i) ( $ {K_f} $ ) (m) $ = {\text{ }}\left( {1.9} \right){\text{ }}\left( {1.86} \right){\text{ }}\left( {0.10125/0.100} \right){\text{ }} = {\text{ }}{3.578^\circ }C $
Hence the freezing point can be calculated as:
$ {T_f} $ = original freezing point – change in freezing point
$
= {\text{ }}0{\text{ }}-{\text{ }}3.578\;\;\;\;\; \\
\; = {\text{ }} - {\text{ }}3.578 \\
$ .
Note
Freezing point depression is the phenomena that describes why adding a solute to a solvent -results in the lowering of the freezing point of the solvent. When a substance starts to freeze, the molecules slow down due to the decreases in temperature, and the intermolecular forces start to take over.
Complete Step-by-Step Answer
Before we move ahead with the solution of this question, let us first understand some basic concepts:
Molality: Molality can be basically explained as the ratio of the number of moles of the solute to the total mass of the solvent. The unit for the mass of the solvent must be ‘kg’.
Molality = $ \dfrac{{number\,\,of\,\,moles\,\,of\,\,solute}}{{mass\,\,of\,\,solvent}} $ =m
The number of moles of HBr can thus be calculated as:
Number of moles = $ \dfrac{{total\,\,mass}}{{molar\,\,mass}} $ = $ \dfrac{{8.1}}{{80}} = 0.10125 $ moles
Molality = m = $ \dfrac{{0.10125}}{{0.100}} $
In aqueous solution, HBr dissociates as follows:
$ HBr \Leftrightarrow {H^ + } + {\mathbf{B}}{r^ - } $
Hence, even though we only have one compound in hand, upon dissociation we will have two ions, two correct this anomaly, we will use Van’t Hoff Factor. The Van’t Hoff factor for this case can be calculated as:
i = 1 + $ \alpha $ (n-1), where $ \alpha $ is the degree of dissociation and n is the number of ions formed
i = 1 + (2-1) (0.9) = 1.9
Now, the change in freezing point can be calculated as:
$ \Delta {T_f} = $ (i) ( $ {K_f} $ ) (m) $ = {\text{ }}\left( {1.9} \right){\text{ }}\left( {1.86} \right){\text{ }}\left( {0.10125/0.100} \right){\text{ }} = {\text{ }}{3.578^\circ }C $
Hence the freezing point can be calculated as:
$ {T_f} $ = original freezing point – change in freezing point
$
= {\text{ }}0{\text{ }}-{\text{ }}3.578\;\;\;\;\; \\
\; = {\text{ }} - {\text{ }}3.578 \\
$ .
Note
Freezing point depression is the phenomena that describes why adding a solute to a solvent -results in the lowering of the freezing point of the solvent. When a substance starts to freeze, the molecules slow down due to the decreases in temperature, and the intermolecular forces start to take over.
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