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

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