Answer: Boiling point of the given solution is [tex]102.048^{o}C[/tex].
Explanation:
Given: Molality = 2.00 m
[tex]k_{b} = 0.512^{o}C[/tex]
Now, equation for dissociation of water is as follows.
[tex]H_{2}O \rightarrow H^{+} + OH^{-}[/tex]
As it is giving 2 ions upon dissociation. So, the value of i = 2.
Formula used to calculate change in temperature is as follows.
[tex]\Delta T = i \times k_{b} \times m[/tex]
where,
i = Van't Hoff factor
[tex]k_{b}[/tex] = molal boiling point elevation constant
m = molality
Substitute the values into above formula as follows.
[tex]\Delta T = i \times k_{b} \times m\\= 2 \times 0.512^{o}C \times 2.00 m\\= 2.048^{o}C[/tex]
As the boiling point of water is [tex]100^{o}C[/tex]. Hence, the boiling point of solution will be as follows.
[tex]\Delta T^{'}_{b} = 100^{o}C + 2.048^{o}C\\= 102.048^{o}C[/tex]
Thus, we can conclude that boiling point of the given solution is [tex]102.048^{o}C[/tex].
