Answer:
127.3° C, (This is not a choice)
Explanation:
This is about the colligative property of boiling point.
ΔT = Kb . m . i
Where:
ΔT = T° boling of solution - T° boiling of pure solvent
Kb = Boiling constant
m = molal (mol/kg)
i = Van't Hoff factor (number of particles dissolved in solution)
Water is not a ionic compound, but we assume that i = 2
H₂O → H⁺ + OH⁻
T° boling of solution - 118.1°C = 0.52°C . m . 2
Mass of solvent = Solvent volume / Solvent density
Mass of solvent = 500 mL / 1.049g/mL → 476.6 g
Mol of water are mass / molar mass
76 g / 18g/m = 4.22 moles
These moles are in 476.6 g
Mol / kg = molal → 4.22 m / 0.4766 kg = 8.85 m
T° boling of solution = 0.52°C . 8.85 m . 2 + 118.1°C = 127.3°C
Answer : The boiling point of a solution is, [tex]142.5^oC[/tex]
Explanation :
Formula used for Elevation in boiling point :
[tex]\Delta T_b=i\times k_b\times m[/tex]
or,
[tex]T_b-T^o_b=i\times k_b\times \frac{w_2\times 1000}{M_2\times w_1}[/tex]
where,
[tex]T_b[/tex] = boiling point of solution = ?
[tex]T^o_b[/tex] = boiling point of acetic acid = [tex]118^oC[/tex]
[tex]k_b[/tex] = boiling point constant of acetic acid = [tex]2.93^oC/m[/tex]
m = molality
i = Van't Hoff factor = 1
[tex]w_2[/tex] = mass of solute (water) = 76 g
[tex]w_1[/tex] = mass of solvent (acetic acid) = [tex]Density\times Volume=1.049g/mL\times 500mL=524.5g[/tex]
[tex]M_2[/tex] = molar mass of solute (water) = 18 g/mol
Now put all the given values in the above formula, we get:
[tex](T_b-118)^oC=1\times (2.93^oC/m)\times \frac{(76g)\times 1000}{18g/mol\times (524.5g)}[/tex]
[tex]T_b=142.5^oC[/tex]
Therefore, the boiling point of a solution is, [tex]142.5^oC[/tex]
