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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.
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