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Titik beku larutan yang dibuat dengan melarutkan 20,5 g suatu zat yang rumus empirisnya (c3h2)n dalam 400 g benzena adalah 4,33 oc. Titik beku benzene murninya adalah 5,48°c. Rumus molekul senyawa tersebut adalah kf 5,12 C/m
Accepted Answer
The molecular formula of the compound is C9H6. Here's how to determine it:
1. Calculate the molality of the solution:
- ΔTf = Kf
molality
- 5.48 °C - 4.33 °C = 5.12 °C/m
molality
- molality = 0.224 m
2. Calculate the moles of solute:
- molality = moles of solute / kg of solvent
- 0.224 mol/kg = moles of solute / 0.4 kg
- moles of solute = 0.0896 mol
3. Calculate the molar mass of the solute:
- molar mass = mass of solute / moles of solute
- molar mass = 20.5 g / 0.0896 mol
- molar mass = 228.6 g/mol
4. Determine the molecular formula:
- The empirical formula mass of (C3H2)n is 3(12.01 g/mol) + 2(1.01 g/mol) = 40.05 g/mol
- Divide the molar mass of the compound by the empirical formula mass: 228.6 g/mol / 40.05 g/mol ≈ 5.7
- Since the ratio is close to 6, the molecular formula is (C3H2)6, which simplifies to C9H6.
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