Respuesta :
Answer:
146,575 grams of ethylene glycol would need to be dissolved in 15 kg of pure water
The boiling point of the solution is 181.95°C.
Explanation:
Mass of ethylene glycol = x
Molar mass of ethylene glycol = 2 × 12 g/mol + 2 × 32 g/mol+ 4 × 1= 62g/mol
Moles of ethylene glycol =[tex]\frac{x}{62 g/mol}[/tex]
Mass of solvent that is water = 15 kg
The molal freezing point depression constant for water [tex]K_f=1.86 K kg/mol[/tex]
Molality of the solution:
[tex]Molality=\frac{\text{Moles of compound}}{\text{Mass of solvent (kg)}}[/tex]
[tex]Molality=m=\frac{x}{62 g/mol\times 15 kg}[/tex]
Depression in freezing point of water =[tex]\Delta T_f=20^oC=293.15 K[/tex]
((T)°C =T+ 273.15 K)
[tex]\Delta T_f=K_f\times m[/tex]
[tex]293.15 K=1.86 K kg/mol\times \frac{x}{62 g/mol\times 15 kg}[/tex]
x = 146,575 g
Boiling point of this solution =[tex]T_b[/tex]
The molal boiling point elevation constant for water[tex]K_b = 0.52 K kg/mol[/tex]
[tex]\Delta T_b=K_b\times m[/tex]
[tex]\Delta T_b=0.52 K kg/mol\times \frac{146,575 g}{62 g/mol\times 15 kg}[/tex]
[tex]\Delta T_b=81.95K[/tex]
Normal boiling point of water is = T = 373.15 K
[tex]\Delta T_b=T_b-T[/tex]
[tex]T_b=\Delta T_b+T=81.95K+373.15 K=455.1 K=181.95^oC[/tex]
The boiling point of the solution is 181.95°C.
