Determine the freezing point of a solution that contains 78.8 g of naphthalene (C10H8, molar mass=128.16g/mol)

Question

dissolved in 722 mL of benzene (d = 0.877 g/mL). Pure benzene has a melting point of 5.50 degrees C and a freezing point depression constant of 4.90 degrees C/m.

I dont even know how to start to solve this. Please help me!

Answer 1

Mass benzene = 0.877 g/mL x 722 mL =  633.2 g => 0.6332 Kg

moles naphthalene = 78.8 g / 128.17 g/mol =  0.615

molality = 0.615 / 0.6332 =0.971

delta T = 4.90 x 0.971 =4.76

freezing point = 5.50 - 4.76 = 0.74 °C