Question

Determine the freezing point of a solution that contains 78.8 g of naphthalene (C10H8, molar mass = 128.16 g/mol) dissolved in 722 mL of benzene (d = 0.877 g/mL). Pure benzene has a melting point of 5.50°C and a freezing point depression constant of 4.90°C/m.0.74°C4.76°C4.17°C1.68°C1.33°C

Answer 1

Answer:

0.74°C

Explanation:

The freezing point depression is a colligative property that can be calculated using the following expression.

ΔTf = Kf . b

where,

ΔTf: depression in the melting point

Kf: freezing point depression constant

b: molality

The mass of the solvent (benzene) is:

$722mL.\frac{0.877g}{mL} =633g=0.633kg$

The moles of solute (naphtalene) are:

$\frac{78.8g}{128.16g/mol} =0.615mol$

The molality of naphtalene is:

$b=\frac{0.615mol}{0.633kg} = 0.972mol/kg$

ΔTf = Kf . b

ΔTf = (4.90°C/m) . 0.972m = 4.76°C

The melting point of the solution is:

5.50°C - 4.76°C = 0.74°C