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Non-electrolyte = A, 10.6 grams, dissolved in solvent B, 740 grams, the boiling point of the solution is higher than the boiling point of the solvent
Pure dissolved in 0.284 ° C. Determine the Kb value of solvent B. Set the molecular weight of A = 106.

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sebassandin

Answer:

2.1 °C/m

Explanation:

Hello, for this exercise, consider the formula:

[tex]T_{solution}-T{solvent}=K_bm_solute[/tex]

Considering that the difference in the temperature is 0.284°C, and the given molality by:

[tex]m_{solute}=\frac{10.6g\frac{1mol}{106g}}{740g*\frac{1kg}{1000g} } \\m_{solute}=0.135m[/tex]

Now, solving for [tex]K_b[/tex], we get:

[tex]K_b=\frac{0.284C}{0.135m}\\K_b=2.1 C/m[/tex]

Best regards.

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