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Chemistry. Elevation in Boiling Point. Question. Boiling point of water at 750 mm H g is 99.63∘C. How much sucrose is to be added to 500 g of water such that it boils at 100∘C. Solution. Verified by Toppr. Given, T b = 100−99.63 =0.37o;Kb = 0.52. Mass of water =500 g. Let x gm of sucrose be added. Molar mass of sucrose = 342 g mol−1.
Solution. Given: Mass of water (W A)= 500 g. Boiling point of pure water at 750 mm Hg (T 0 b) = 99.630C. Boiling point of mixture (T b) =1000C. Molar mass (M B) of sucrose (C12H22O11) = 12×12+1×22+16×11. = 342 g mol−1. Elevation in boiling point (ΔT b) = T b−T 0 b. ΔT b= 100−99.63= 0.370C or K. Amount of sucrose.
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Dec 13, 2017 · Best answer. Here, elevation of boiling point ∆Tb = (100 + 273) − (99.63 + 273) = 0.37 K. Mass of water, wl = 500 g. Molar mass of sucrose (C12H22O11), M2 = 11 × 12 + 22 × 1 + 11 × 16. = 342 g mol−1. Molal elevation constant, Kb = 0.52 K kg mol−1 We know that: = 121.67 g (approximately) Hence, 121.67 g of sucrose is to be added.
Answers (1) Here we will use the formula : Elevation in temperature = 100 - 99.63 = 0.37. K b = 0.52 ; Putting all values in above formula, we get : Thus 121.67 g of sucrose needs to be added. Posted by.
