Respuesta :
To answer the question above, solve first for the boiling point elevation through the equation,
ΔT = (Kb)(m)
where ΔT is the boiling point, Kb is constant for water, and m is the concentration. Substituting the known values,
ΔT = (0.51 K/molal)(0.6 molal)
ΔT = 0.306 K
The boiling point therefore of the solution is equal to 100.306°C.
Answer: 100.306°C
ΔT = (Kb)(m)
where ΔT is the boiling point, Kb is constant for water, and m is the concentration. Substituting the known values,
ΔT = (0.51 K/molal)(0.6 molal)
ΔT = 0.306 K
The boiling point therefore of the solution is equal to 100.306°C.
Answer: 100.306°C
Boiling point of 0.6 molar solution of NaCl in water is [tex]\boxed{-172.844^\circ\text{C}}[/tex].
Further Explanation:
Colligative properties:
These properties are dependent only on concentration of solute particles, no matter what type of solute is used.
Following four are colligative properties:
- Relative lowering of vapor pressure.
- Elevation in boiling point.
- Depression in freezing point.
- Osmotic pressure.
Boiling point elevation is evaluated by following formula:
[tex]\Delta\text{T}_\text{b}=\text{k}_\text{b}\text{m}[/tex] ...... (1)
Here,
[tex]\Delta\text{T}_\text{b}[/tex] is elevation in boiling point.
[tex]\Delta\text{k}_\text{b}[/tex] is molal boiling point constant.
m is molality of solution.
Substitute 0.6 m for m and 0.51 K/m for [tex]\Delta\text{k}_\text{b}[/tex] in equation (1).
[tex]\begin{aligned}\Delta\text{T}_\text{b}=&\text{(0.51 K/m)}(0.6\text{m})\\=&0.306\text{K}\end{aligned}[/tex]
Boiling point elevation is converted from Kelvin to degrees Celsius with help of following conversion factor:
[tex]0^\circ\text{C}=273.15\text{K}[/tex]
Therefore boiling point elevation is calculated as follows:
[tex]\begin{aligned}\Delta\text{T}_\text{b}&=\text{0.306 K}-\text{273.15 K}\\&=-272.844^\circ\text{C}\end{aligned}[/tex]
The formula to calculate the temperature change in the NaCl solution is as follows:
[tex]\Delta\text{T}_\text{b}=\text{Boiling point of NaCl}-\text{Boiling point of H}_{2}\text{O}[/tex] ...... (2)
Rearrange equation (2) to calculate boiling point of NaCl.
[tex]\text{Boiling point of NaCl}=\Delta\text{T}_\text{b}+\text{Boiling point of H}_{2}\text{O}[/tex] ....... (3)
Substitute [tex]100^\circ\text{C}[/tex] for boiling point of pure [tex]\text{H}_{2}\text{O}[/tex] and [tex]-272.844^\circ\text{C}[/tex] for [tex]\Delta\text{T}_\text{b}[/tex] in equation (3).
[tex]\begin{aligned}\text{Boiling point of NaCl}=&-272.844^\circ\text{C}+100^\circ\text{C}\\=&-172.844^\circ\text{C}\end{aligned}[/tex]
Learn more:
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Answer details:
Grade: Senior School
Chapter: Colligative properties
Subject: Chemistry
Keywords: colligative properties, boiling point, elevation in boiling point, boiling point of NaCl, NaCl, concentration, solute particles, relative lowering of vapor pressure, depression in freezing point, osmotic pressure, -172.844 degrees Celsius.
