A certain substance X melts at a temperature of 5.8°C. But if a 550.g sample of X is prepared with 30.75g of urea NH22CO dissolved in it, the sample is found to have a melting point of −1.6°C instead. Calculate the molal freezing point depression constant Kf of X.Round your answer to 2 significant digits.
Question
A certain substance X melts at a temperature of 5.8°C. But if a 550.g sample of X is prepared with 30.75g of urea NH22CO dissolved in it, the sample is found to have a melting point of −1.6°C instead. Calculate the molal freezing point depression constant Kf of X.Round your answer to 2 significant digits.
Solution
To solve this problem, we need to use the formula for freezing point depression:
ΔTf = Kf * m
where:
- ΔTf is the change in freezing point (original freezing point - new freezing point)
- Kf is the freezing point depression constant we're trying to find
- m is the molality of the solution (moles of solute/kg of solvent)
Step 1: Calculate ΔTf ΔTf = original freezing point - new freezing point ΔTf = 5.8°C - (-1.6°C) = 7.4°C
Step 2: Calculate the molality (m) First, we need to find the moles of urea. The molar mass of urea (NH22CO) is approximately 60.06 g/mol. moles of urea = mass/molar mass = 30.75g / 60.06 g/mol = 0.512 mol
Next, we need to convert the mass of the solvent (substance X) from grams to kilograms: 550.g = 0.55 kg
Now we can calculate the molality: m = moles of solute/kg of solvent = 0.512 mol / 0.55 kg = 0.931 mol/kg
Step 3: Solve for Kf Now we can rearrange the freezing point depression formula to solve for Kf: Kf = ΔTf / m = 7.4°C / 0.931 mol/kg = 7.95 °C·kg/mol
So, the molal freezing point depression constant Kf of substance X is approximately 7.95 °C·kg/mol, rounded to two significant digits.
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