Question

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

Answer 1

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 2

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

Explanation :

Formula used for Elevation in boiling point :

$\Delta T_b=i\times k_b\times m$

or,

$T_b-T^o_b=i\times k_b\times \frac{w_2\times 1000}{M_2\times w_1}$

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) = $Density\times Volume=1.049g/mL\times 500mL=524.5g$

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

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

$(T_b-118)^oC=1\times (2.93^oC/m)\times \frac{(76g)\times 1000}{18g/mol\times (524.5g)}$

$T_b=142.5^oC$

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