The freezing point of camphor is 178.4�c and its molal freezing point depression constant (kf) is 37.7�c�kg/mol. how many grams of naphthalene (c10h8; molar mass = 128.17 g/mol; a solute, not the solvent, in this question) should be mixed with 12.0 g of camphor to lower the freezing point to 170.0 �c?

Answer :

The correct answer is : 0.34 g of naphthalene .

Step 1 : To find molaity of solution using freezing point depression formula .

The formula for Freezing point depression is :ΔTf = kf * m

where , ΔTf = Freezing point of solvent - freezing point of solution

kf = freezing point depression constant and m = molality .

Given : Freezing point of solvent (camphor ) = 178.4 °C

Freezing point of solution ( naphthalene in camphor) = 170.0°C

kf for camphor = 37.7 °C Kg /mol

Plugging values in ΔTf - kf *m

178.4 °C - 170.0°C = 37.7 °C Kg/mol * m

8.4 °C = 37.7 °C Kg/mol * m

Dividing both side by 37.7 °C Kg/mol to isolate m

8.4 °C /37.7°C Kg/mol = 37.7 °C Kg/mol / 37.7 °C Kg/mol * m

m (molality ) = 0.22 mol/Kg

Step 2 : To find moles from molality .

Molality is moles of naphthalene ( solute ) present in kilograms of camphor (solvent ) . Formula of molality is : moles of solute / kilograms of solvent

Given mass of camphor = 12 g or 0.012 Kg ( 1 g = 0.001 Kg )

Plugging value in formula => 0.22 mol /Kg = mole of solute / 0.012Kg

mole of solute = 0.22 mol/Kg x 0.012 Kg = 0.00264 mol

Step 3 : To convert mole of naphthalene to its mass :

mass = mole x molar mass

Mass = 0.00264 mol x 128.17 g/mol

Mass of naphthalene to be added = 0.34 g

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