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Assuming ideal solution behavior, what is the boiling point of a solution of 115.0 g of nonvolatile sucrose (table sugar), C₁₂H₂₂O₁₁ (342.300 g/mol), in 350.0 g of water (Kb = 0.512 °C m⁻¹; boiling point = 100.0 °C)?

Sagot :

tushargupta0691

The boiling point of the solution will be 100.97 ° C.

[tex]T_f[/tex], pure water = 0.00 ° Celsius.

[tex]T_b[/tex] , pure water = 100° Celcius

[tex]K_b[/tex] = 0.512 C kg/mol

[tex]K_f[/tex] = -1.86 C kg/mol.

Given,

[tex]K_b = \frac{0.512 C kg}{mol}[/tex]

[tex]m = \frac{115g \frac{mol }{342.300g} }{0.35 kg}[/tex]

m = 0.95

i = 2 ( assuming no ion pairing)

Now, recall,

Δ[tex]T_b[/tex] =[tex]iK_bm[/tex]

Hence,

Δ[tex]T_b[/tex] = [tex]iK_bm[/tex]

      = 2 × 0.512 × 0.95

      = 0.97 °C

Therefore, the boiling point of the solution will be 100.97 °C.

Learn more about the solution here:

https://brainly.com/question/1597618

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