Question

How much anthracene (MW = 178.23 g/mol) would need to be added to 100.0 g of cyclohexane (Kfp = 20.4°C/m) to depress the melting point by 5.50°C. Anthracene is a non-electrolyte.

Answer 1

This problem is asking for the mass of anthracene one should use to decrease the melting point of a solution with 100.0 g of cyclohexane, and its freezing point depression constant, by 5.50 °C. At the end, the result turns out to be 4.81 g.

Freezing point depression:

In chemistry, colligative properties are the result from mixing a solute and a solvent in order to modify the properties of the latter due to the addition of the former. Thus, freezing point depression, boiling point elevation, osmotic pressure and vapor pressure lowering are colligative properties one can go over.

In this case, we talk about freezing point depression, or even melting, because anthracene is added to cyclohexane to decrease its freezing, or melting, point by 5.50 °C. In such a way, we can calculate the mass of anthracene by firstly calculating the molality of the solution with:

$\Delta T_f=-m*Kf\\\\m=\frac{-\Delta T_f}{Kf} \\\\m=\frac{-(-5.50\°C)}{20.4\°C/m}\\\\m=0.270mol/kg$

Then, as 100.0 g of cyclohexane are equivalent to 0.1000 kg and, given the molar mass of anthracene, we can calculate the mass with the previous molality:

$m_{anthracene}=0.270\frac{mol}{kg}*0.1000kg*\frac{178.23g}{1mol} \\\\m_{anthracene}=4.81g$

Learn more about colligative properties: https://brainly.com/question/10323760