Respuesta :
Answer:
∆T = imK
∆T = Change in boiling point (B.P.)
i = van't Hoff factor = 2 for NaCl (Na+ and Cl2_
m = molality = 0.321 m
k = boiling point constant for water = 0.512 deg/m
∆T = (2)(0.321)(0.512) = 0.329 degrees
Since the normal B.P. for water is 100ºC, the new boiling point of this solution is 100 + 0.329 = 100.329ºC
So Our Answer is 100.329ºC
The boiling point of the 0.321 m NaCl solution has been 100.329 degrees Celsius.
The boiling point has been defined at the temperature at which the liquid started to convert to gas.
The van't Hoff factor has been a value of the number of ions formed by the dissociation of 1 formula unit of a compound.
The change in the boiling point of the sample, ([tex]\Delta T[/tex]) has been given by:
[tex]\Delta T=imK[/tex]
Where, the van't Hoff factor for NaCl, i =-2
The molality of the sample, m =0.321 m
The boiling point constant of water, [tex]K=0.512\;^\circ m^-^1[/tex]
Substituting the values for change in temperature ([tex]\Delta T[/tex]):
[tex]\Delta T=2\;\times\;0.321\;\times\;0.512\;^\circ C\\\Delta T=0.329\;^\circ C[/tex]
The change in the temperature of the water on the addition of NaCl has been 0.329 degrees Celsius.
The initial temperature of the water has been 100 degrees Celsius. The new temperature (T) of the NaCl solution has been:
[tex]T=100\;+\;0.329\;^\circ \text C\\T=100.329\;^\circ \text C\\[/tex]
The new boiling point of the NaCl solution has been 100.329 degrees Celsius.
For more information about boiling point, refer to the link:
https://brainly.com/question/2153588
