Question

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.

Answer 1

Answer:

2.1 °C/m

Explanation:

Hello, for this exercise, consider the formula:

$T_{solution}-T{solvent}=K_bm_solute$

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

$m_{solute}=\frac{10.6g\frac{1mol}{106g}}{740g*\frac{1kg}{1000g} } \\m_{solute}=0.135m$

Now, solving for $K_b$, we get:

$K_b=\frac{0.284C}{0.135m}\\K_b=2.1 C/m$

Best regards.