Question

Calculate the freezing point and melting point of a solution containing 10.0 g of naphthalene (C₁₀H₈) in 100.0 mL of benzene. Benzene has a density of 0.877 g/cm³

Answer 1

Remember the equation for the depression of the freezing point:

ΔTf = K_f * m (1)

Where:

K_f is a constant. For benzene, K_f = 5.12

m is for molality (mols of solute / kg of solvent)

Step 1:

Whe need the moles of naphthalene, so we use the molecular mass which is 128 g/mol

moles of C10H8 = 10 g / 128 g/mol = 0.078 moles

Step 2:

Also, we need the mass of benzene. Remember 1 ml = 1cm³

100 ml benzene * 0.877 g/ml * 1 kg/1000 g = 0.0877 kg

Step 3:

molality = 0.078 moles / 0.0877 kg = 0.89 mol/kg

Step 4:

Now, let's use equation (1)

ΔTf = K_f *m = -5.12 °C/m * 0.89 m = - 4.6 °C

Step 5:

The original freezing point of benzene is 5,5 °C, so:

ΔTf = T_{f solvent} - T_fsolution

4.6 °C = 5.5 - x

T_{f solution} = 5.5 - 4.6 °C = 0.9 °C

Also, you need to realize that the freezing point and the melting point are just two faces of the same coin. If you have a solid and you want to melt it, you heat up the solid until it surpasses the freezing point (in this case, over 0.9 °C). If you want to freeze it, you cold it under 0.9°C.