Question

Calculate the freezing point (in degrees C) of a solution made by dissolving 7.99 g of anthracene{C14H10} in 79 g of benzene. The Kfp of the solvent is 5.12 K/m and the normal freezing point is 5.5 degrees C. Enter your answer with 2 decimal places.

Answer 1

To determine the freezing point of a solution made with anthracene dissolved in benzene, the molality is first calculated, then the freezing point depression is determined using the freezing point depression constant of benzene, and finally, it is subtracted from the normal freezing point of benzene. The calculated freezing point of this solution is 1.90 °C.

Step-by-step explanation:

To calculate the freezing point of a solution made by dissolving 7.99 g of anthracene (C14H10) in 79 g of benzene, we first need to determine the molar mass of anthracene. From given information, the molar mass of anthracene is 144 g mol-1. Next, we calculate the number of moles of anthracene dissolved:

Number of moles = mass / molar mass = 7.99 g / 144 g mol-1 = 0.0555 mol

Now, we calculate the molality (m) of the solution:

Molality (m) = moles of solute / kg of solvent = 0.0555 mol / 0.079 kg = 0.703 mol kg-1

Then, we use the formula for freezing point depression:

ΔTf = Kfp * m

ΔTf = 5.12 K/m * 0.703 mol kg-1 = 3.60 K

Finally, we subtract the calculated freezing point depression from the normal freezing point of benzene to find the freezing point of the solution:

Freezing point of solution = normal freezing point - ΔTf = 5.5 °C - 3.60 °C

The freezing point of the solution is therefore 1.90 °C.

Answer 2

Answer: The freezing point of solution is 2.6°C

Step-by-step explanation:

To calculate the depression in freezing point, we use the equation:

`\Delta T_f=iK_fm`

Or,

`\Delta T_f=i* K_f* \frac{m_(solute)* 1000}{M_(solute)* W_(solvent)\text{ in grams}}`

where,

`\Delta T_f` =
`\text{Freezing point of pure solution}-\text{Freezing point of solution}`

Freezing point of pure solution = 5.5°C

i = Vant hoff factor = 1 (For non-electrolytes)

`K_f` = molal freezing point depression constant = 5.12 K/m = 5.12 °C/m

`m_(solute)` = Given mass of solute (anthracene) = 7.99 g

`M_(solute)` = Molar mass of solute (anthracene) = 178.23 g/mol

`W_(solvent)` = Mass of solvent (benzene) = 79 g

Putting values in above equation, we get:

`5.5-\text{Freezing point of solution}=1* 5.12^oC/m* (7.99* 1000)/(178.23g/mol* 79)\\\\\text{Freezing point of solution}=2.6^oC`

Hence, the freezing point of solution is 2.6°C