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

2.1 °C/m

Step-by-step 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)=(10.6g(1mol)/(106g))/(740g*(1kg)/(1000g) ) \\m_(solute)=0.135m`

Now, solving for
`K_b`, we get:

`K_b=(0.284C)/(0.135m)\\K_b=2.1 C/m`

Best regards.