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
