Respuesta :
Answer:The boiling point of the solution is 108° C.
Explanation:
Boiling point of pure water=T=[tex]100^oC[/tex]
Boiling point of water after addition of 500 g of ethylene glycol=[tex]T_f[/tex]
Mass of water = 500g = 0.5 kg (1000 g = 1 kg)
[tex]\Delta T_f=K_b\times \frac{\text{Mass of ethlyene glycol}}{\text{Molar mass of ethylene glycol}}[/tex]
[tex]\Delta T_f=0.512^oC/m\times \frac{500 g}{62.07 g/mol\times 0.5 kg}[/tex]
[tex]\Delta T_f=8.24 ^oC[/tex]
[tex]\Delta T_f=T_f-T[/tex]
[tex]8.24^oC=T_f-100^oC[/tex]
[tex]T_f=108.24^oC[/tex]
The boiling point of the solution is 108° C.
Boiling point is defined as the temperature at which vapor pressure of a substance is equal to the surrounding pressure. The boiling point of solution will be 108-degree celcius.
Given that,
Boiling point of water = 100-degree celcius
Boiling point after the addition of 500 g of ethylene glycol = [tex]{_f}[/tex][tex]\text T_f[/tex]
Mass of water = 500g = 0.5 kg (1000 g = 1 kg)
Now, using the formula:
[tex]\Delta \text{T}_f&=\text {K}_b\dfrac{\text {Mass of ethylene glycol}}{\text{Molar Mass of ethylene glycol}}\\\\\Delta \text{T}_f&= {0.512}^{0}\text C \times\dfrac {500}{62.07 \text{g/mol} \times\; 0.5 \text{kg}}}\\\\\Delta \text T& = 8.24^{0}\\\\\Delta \text T_f = \text T_f - \text T\\\\\text T_f = 108.24^{0}[/tex]
Therefore, the boiling point of water is 108-degree celcius.
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https://brainly.com/question/12972840?referrer=searchResults
