Question

What is the boiling point of a solution of 76 g of water dissolved in 500 mL of acetic acid, CH3COOH?

The density of CH3COOH is 1.049 g/cm3
123.09°C

99.0°C

119.13°C

100.13°C

142.5°C

Answer 1

127.3° C, (This is not a choice)

Step-by-step 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 2

Answer : The boiling point of a solution is,
`142.5^oC`

Explanation :

Formula used for Elevation in boiling point :

`\Delta T_b=i* k_b* m`

or,

`T_b-T^o_b=i* k_b* (w_2* 1000)/(M_2* w_1)`

where,

`T_b` = boiling point of solution = ?

`T^o_b` = boiling point of acetic acid =
`118^oC`

`k_b` = boiling point constant of acetic acid =
`2.93^oC/m`

m = molality

i = Van't Hoff factor = 1

`w_2` = mass of solute (water) = 76 g

`w_1` = mass of solvent (acetic acid) =
`Density* Volume=1.049g/mL* 500mL=524.5g`

`M_2` = molar mass of solute (water) = 18 g/mol

Now put all the given values in the above formula, we get:

`(T_b-118)^oC=1* (2.93^oC/m)* ((76g)* 1000)/(18g/mol* (524.5g))`

`T_b=142.5^oC`

Therefore, the boiling point of a solution is,
`142.5^oC`