Answer :
Answer : The boiling point of a solution is, [tex]100.42^oC[/tex]
Explanation :
Formula used for Elevation in boiling point :
[tex]\Delta T_b=k_b\times m[/tex]
or,
[tex]T_b-T^o_b=\frac{1000\times k_b\times w_2}{w_1\times M_2}[/tex]
where,
[tex]T_b[/tex] = boiling point of solution = ?
[tex]T^o_b[/tex] = boiling point of pure water = [tex]100^oC[/tex]
[tex]k_b[/tex] = boiling point constant for water = [tex]0.512^oC/m[/tex]
m = molality
[tex]w_2[/tex] = mass of solute (sucrose) = 4.27 g
[tex]w_1[/tex] = mass of solvent (water) = 15.2 g
[tex]M_2[/tex] = molar mass of solute (sucrose) = 342.3 g/mole
Now put all the given values in the above formula, we get the boiling point of a solution.
[tex]T_b-100^oC=\frac{1000\times 0.512^oC/m\times 4.27g}{15.2g\times 342.3g/mole}[/tex]
[tex]T_b=100.42^oC[/tex]
Therefore, the boiling point of a solution is, [tex]100.42^oC[/tex]
The temperature of the solution is 100.42°C.
Given that;
ΔT = K m i
Where;
ΔT = boiling point elevation
K = boiling point constant for water
m = molality of the solution
i = Van't Hoff factor
Number of moles of solute = 4.27 g /342 g/mol = 0.0125 moles
Molality of the solution = 0.0125 moles/15.2 × 10^-3 Kg = 0.822 m
Since the boiling point of pure water = 100°C
Let the boiling point of pure water be Ta
Let the boiling point of the solution be Tb
ΔT = Tb - Ta = Tb - 100
Substituting values;
Tb - 100 = 0.512c/m × 0.822 m × 1
Note that the Van't Hoff factor (i) = 1 because the solute is molecular
Tb = [0.512°C/m × 0.822 m × 1] + 100°C
Tb = 100.42°C
Learn more: https://brainly.com/question/15178305