Question

The freezing point of benzene is 5.5°C. What is the freezing point of a solution of 2.80 g of naphthalene (C10H8) in 355 g of benzene (Kf of benzene = 4.90°C/m)?

Please answer in how many degrees of Celcius

Answer 1

Answer – We are given, mass of benzene = 355 g ,mass of naphthalene = 2.80 g

The freezing point of benzene = 5.5°C, Kf of benzene = 4.90°C/m

First we need to calculate the molality of naphthalene

We know,

Moles of naphthalene = 2.8 g / 128.17 g.mol^-1

                                   = 0.0218 moles

Molality of naphthalene = 0.0218 moles / 0.355 kg

                                       = 0.0615 m

Now we know the freezing point depression formula

∆Tf = Kf*m

       = 4.90°C/m * 0.0615 m

       = 0.302 °C

So, freezing point of solution = freezing point of pure solvent - ∆Tf

                                                = 5.5°C - 0.302 °C

                                                = 5.2°C