
The correct relationship between the boiling points of very dilute solutions of \[AlC{l_3}\] \[\left( {{T_1}} \right)\] and \[CaC{l_2}\,\,\left( {{T_2}} \right)\] , having the same molar concentration is:
A. \[\;{T_1} = {T_2}\]
B. ${T_1} > {T_2}$
C. ${T_2} > {T_1}$
D. ${T_2} \geqslant {T_1}$
Answer
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Hint: The temperature at which a liquid's vapour pressure equals the pressure around it and the liquid transforms into a vapour is known as the boiling point. A liquid's boiling point varies depending on the atmospheric pressure in the area and its vapour pressure.
Complete step-by-step answer: At boiling temperature, the addition of heat causes the liquid to turn into its vapour without raising the temperature.
The ratio of a substance's mass concentration to the concentration of the particles that are produced when it dissolves is known as the Van't Hoff factor.
We know that boiling point of a solution, ${T_{bs}}$ , is sum of boiling point in pure form, ${T_b}$ , and elevation in boiling point, $\vartriangle {T_b}$ . Also, elevation in boiling point, $\vartriangle {T_b} = {K_b} \times m \times i$ , where $m$ is molality and $i$ is the Van’t Hoff factor, ${K_b}$ is a constant.
This implies that higher the $i$ factor greater will be the elevation of boiling point. Thus, the boiling point of solution also increases.
Now, we know that $AlC{l_3}$ dissociates as follows:
$AlC{l_3}\,\, \rightleftharpoons \,\,A{l^{3 + }}\, + \,3C{l^ - }$
This implies $i = 4$ for $AlC{l_3}$ . ...(1)
Also, $CaC{l_2}$ dissociates as:
$CaC{l_2}\,\, \rightleftharpoons \,\,C{a^{2 + }}\, + \,2C{l^ - }$
This implies that $i = 3$ for $CaC{l_2}$. ...(2)
From equations (1) and (2), we see that the value of the Van't Hoff factor is greater for $AlC{l_3}$ . This implies that it has a higher elevation of boiling point. Hence, $AlC{l_3}$ has a higher boiling point of the solution.
From this we get, ${T_1} > {T_2}$ .
Option ‘B’ is correct
Note: A solution has a greater boiling point than a pure solvent because of a phenomena known as boiling-point elevation, which occurs when one chemical is added to a liquid to raise its boiling point. When a pure solvent, like water, is combined with a non-volatile solute, like salt, this occurs.
Complete step-by-step answer: At boiling temperature, the addition of heat causes the liquid to turn into its vapour without raising the temperature.
The ratio of a substance's mass concentration to the concentration of the particles that are produced when it dissolves is known as the Van't Hoff factor.
We know that boiling point of a solution, ${T_{bs}}$ , is sum of boiling point in pure form, ${T_b}$ , and elevation in boiling point, $\vartriangle {T_b}$ . Also, elevation in boiling point, $\vartriangle {T_b} = {K_b} \times m \times i$ , where $m$ is molality and $i$ is the Van’t Hoff factor, ${K_b}$ is a constant.
This implies that higher the $i$ factor greater will be the elevation of boiling point. Thus, the boiling point of solution also increases.
Now, we know that $AlC{l_3}$ dissociates as follows:
$AlC{l_3}\,\, \rightleftharpoons \,\,A{l^{3 + }}\, + \,3C{l^ - }$
This implies $i = 4$ for $AlC{l_3}$ . ...(1)
Also, $CaC{l_2}$ dissociates as:
$CaC{l_2}\,\, \rightleftharpoons \,\,C{a^{2 + }}\, + \,2C{l^ - }$
This implies that $i = 3$ for $CaC{l_2}$. ...(2)
From equations (1) and (2), we see that the value of the Van't Hoff factor is greater for $AlC{l_3}$ . This implies that it has a higher elevation of boiling point. Hence, $AlC{l_3}$ has a higher boiling point of the solution.
From this we get, ${T_1} > {T_2}$ .
Option ‘B’ is correct
Note: A solution has a greater boiling point than a pure solvent because of a phenomena known as boiling-point elevation, which occurs when one chemical is added to a liquid to raise its boiling point. When a pure solvent, like water, is combined with a non-volatile solute, like salt, this occurs.
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