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Questions
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.
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?
Solution
Here, elevation of boiling point ΔTb = (100 + 273) − (99.63 + 273)
= 0.37 K
Mass of water, w1 = 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:
`triangle"T"_"b" = ("K"_"b"xx1000xx"w"_2)/("M"_2xx"w"_1)`
`=>"w"_2 = (triangle"T"_"b"xx"M"_2xx"w"_1)/("K"_"b"xx1000)`
`= (0.37xx342xx500)/(0.52xx1000)`
= 121.67 g (approximately)
= 122 g
Hence, 122 g of sucrose is to be added.
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