I'd say he ways about 35 kilograms, but I'm probably wrong, xD
i believe
2 g/mL
good luck!
Answer: 120g/mol
Explanation:
The first step we are to take is to calculate the freezing point depression of the solution.
ΔT(f) = freezing point of pure solvent - freezing point of solution
ΔT(f) = 5.48 - 3.77
ΔT(f) = 1.71°C
Next we are to calculate the molal concentration of the solution using freezing point depression
ΔT(f) = K(f) * m
m = ΔT(f)/K(f)
m = 1.71/5.12
m = 0.333 molal
Now, we calculate the molecular weight of the unknown...
m = 0.333 mol = 0.333 mol X per kg of benzene
moles of X = 0.333 mol of X per kg of benzene * 0.5kg of benzene
moles of X = 0.1665
molecular weight of X = 20g of X/0.1665
molecular weight of X = 120/mol
