
What is the boiling point for a \[0.743m\] aqueous solution of \[KCl\] ?
Answer
511.5k+ views
Hint: The temperature at which a liquid's vapour pressure equals the pressure surrounding the liquid, and the liquid turns into a vapour is known as the boiling point of a material. The surrounding air pressure has an effect on a liquid's boiling point.
Complete answer:
We are using the formula,
\[\Delta {T_b} = {T_b} - T_b^ * = i{K_b}m\]
Where,
\[\Delta {T_b}\] is the change in boiling point in \[^\circ C\] from the pure solvent \[T_b^ * \] to the solution \[{T_b}\] .
\[i\] is the Van't Hoff factor, or an effective number of solute particles in solution. The Van't Hoff factor is ratio of the observed colligative property to the calculated colligative property.
\[{K_b} = 0.512^\circ C/m\] is the boiling point constant of the water.
\[m\] is the molality of the solution.
Here we are assuming \[100\% \] dissociation of \[KCl\] .
The chemical reaction equation of \[KCl\] is given as:
\[KCl\left( {aq} \right) \to {K^ + }\left( {aq} \right) + C{l^ - }\left( {aq} \right)\]
We can find that the effective number of solute particles \[i \approx 1 + 1\] .
Therefore,
\[{T_b} - T_b^ * = i{K_b}m\]
That is, \[{T_b} = T_b^ * + i{K_b}m\]
After substituting the values, we got
\[{T_b} = 100^\circ C + 2 \times 0.512^\circ C/m \times 0.743m\]
\[{T_b} = 100.761^\circ C\]
Hence, the boiling point of the aqueous solution is \[100.761^\circ C\] .
Additional Information:
Remember the formula, \[\Delta {T_b} = {T_b} - T_b^ * = i{K_b}m\] . Keep in mind that when another compound is applied to a liquid, that is, a solvent, the boiling point of the liquid increases, implying that a solution has a higher boiling point than a pure solvent.
Note:
The Van 't Hoff factor, \[i\] is a measurement of a solute's effect on colligative properties like osmotic pressure, relative vapour pressure reduction, boiling-point elevation, and freezing-point depression. The ratio between the real concentration of particles formed when a substance is dissolved, and the concentration of a substance measured from its mass is known as the Van 't Hoff factor.
Complete answer:
We are using the formula,
\[\Delta {T_b} = {T_b} - T_b^ * = i{K_b}m\]
Where,
\[\Delta {T_b}\] is the change in boiling point in \[^\circ C\] from the pure solvent \[T_b^ * \] to the solution \[{T_b}\] .
\[i\] is the Van't Hoff factor, or an effective number of solute particles in solution. The Van't Hoff factor is ratio of the observed colligative property to the calculated colligative property.
\[{K_b} = 0.512^\circ C/m\] is the boiling point constant of the water.
\[m\] is the molality of the solution.
Here we are assuming \[100\% \] dissociation of \[KCl\] .
The chemical reaction equation of \[KCl\] is given as:
\[KCl\left( {aq} \right) \to {K^ + }\left( {aq} \right) + C{l^ - }\left( {aq} \right)\]
We can find that the effective number of solute particles \[i \approx 1 + 1\] .
Therefore,
\[{T_b} - T_b^ * = i{K_b}m\]
That is, \[{T_b} = T_b^ * + i{K_b}m\]
After substituting the values, we got
\[{T_b} = 100^\circ C + 2 \times 0.512^\circ C/m \times 0.743m\]
\[{T_b} = 100.761^\circ C\]
Hence, the boiling point of the aqueous solution is \[100.761^\circ C\] .
Additional Information:
Remember the formula, \[\Delta {T_b} = {T_b} - T_b^ * = i{K_b}m\] . Keep in mind that when another compound is applied to a liquid, that is, a solvent, the boiling point of the liquid increases, implying that a solution has a higher boiling point than a pure solvent.
Note:
The Van 't Hoff factor, \[i\] is a measurement of a solute's effect on colligative properties like osmotic pressure, relative vapour pressure reduction, boiling-point elevation, and freezing-point depression. The ratio between the real concentration of particles formed when a substance is dissolved, and the concentration of a substance measured from its mass is known as the Van 't Hoff factor.
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