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Boiling point of water at 750 mm Hg is 99.63°C. How much sucrose is to be added to 500 g of water such that it boils at 100°C. Molal elevation constant for water is 0.52 K kg mol⁻¹

User JChan
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Final answer:

To raise the boiling point of 500 g of water to 100°C, 121.41 g of sucrose should be added. This calculation uses the molal elevation constant for water and the molar mass of sucrose.

Step-by-step explanation:

To calculate how much sucrose must be added to 500 g of water to elevate its boiling point to 100°C, we utilize the concept of boiling point elevation. The formula is ΔT = i * Kb * m, where ΔT is the change in boiling point, i is the van't Hoff factor (which is 1 for sucrose as it does not dissociate), Kb is the molal elevation constant for water, and m is the molality of the solution.

First, we need to establish the change in boiling point (ΔT). Given that the boiling point of water at 750 mm Hg is 99.63°C and we want to achieve 100°C, ΔT = 100°C - 99.63°C = 0.37°C. Utilizing the Kb value of 0.52 K kg mol−1, we rearrange the formula to solve for molality (m = ΔT / (i * Kb)).

m = 0.37°C / (1 * 0.52 K kg mol−1) = 0.71 mol/kg. Since we have 500 g of water, which is 0.5 kg, the amount in moles of sucrose needed is 0.71 mol/kg * 0.5 kg = 0.355 mol.

The molar mass of sucrose (C12H22O11) is approximately 342 g/mol. Now, we can determine the mass of sucrose required: Mass = molar mass * number of moles = 342 g/mol * 0.355 mol = 121.41 g.

Therefore, approximately 121.41 g of sucrose needs to be added to 500 g of water to increase its boiling point to 100°C.

User Divan
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