Chemistry help..!!?

Question

Q part1: A solution of a hydrocarbon dissolved in benzene freezes at -0.51(degree celcius). What is the molality of this solution?

Q part2: If the solution in (Q part1) was made by dissolving 4.367gram of the hydrocarbon in 21.35gram of benzene, how many mole of hydrocarbon were dissolved in the 21.35gram of benzene? + what is the molar mass of hydrocarbon?

please help me.. thanks =D

Answer 1

Q part1:
Melting point of benzene: 5.5C
Kf of Benzene: 5.12 K·kg /mol.
Molality of this solution: (5.5C + 0.51C)/(5.12 K·kg /mol) = 1.2mol/Kg.
Q part2:
# of mole of hydrocarbon in the 21.35gram of benzene:
(1.2mol/Kg)*(21.35gram) = 0.025 mol
the molar mass of hydrocarbon:
(4.367gram)/ (0.025 mol) = 174 g/mol