The freezing point of benzene is 5.5°C. What is the freezing point of a solution of 5.00 g of naphthalene (C10H8) in 344 g benzene? (Kf of benzene = 4.90°C/m.)

4.94°C, the temperature for freezing the solution

Explanation:

Freezing point depression to solve this.

Formula = T° freezing pure solvent - T° freezing solution = Kf . m

With the data given, let's determine m (molality)

Molality → mol/kg (moles of solute in 1kg of solvent)

We need to convert the 344 g to kg → 344 g . 1kg/1000 g = 0.344 kg

Let's determine the moles of solute (naphtalene)

5 g / 128 g/mol = 0.039 mol

Molality → 0.039 mol / 0.344 kg → 0.113

Let's go back to the formula:

5.5°C - T° freezing of solution = 4.90°C /m. 0.113 m

T° freezing of solution = - ( 4.90°C /m. 0.113 m - 5.5°C)

T° freezing of solution =  4.94 °C

0.88m hope i am correct : )