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The vapour pressure of water at 293K is 17.535mmhg.Calculate the vapour pressure of water at 293K when 25g of glucose is dissolved in 450g of water.

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7 months ago Saurabh Koranglekar
10233 Points
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7 months ago
```							Vapour pressure of water, p1° = 17.535 mm of Hg Mass of glucose, w2 = 25 g Mass of water, w1 = 450 g We know that, Molar mass of glucose (C6H12O6), M2 = 6 × 12 + 12 × 1 + 6 × 16 = 180 g mol - 1 Molar mass of water, M1 = 18 g mol - 1 Then, number of moles of glucose, n1 = 25/180 = 0.139 mol And, number of moles of water, n2 =450/18 = 25 mol Now, we know that, (p1° - p°) / p1° = n1 / n2 + n1 ⇒ 17.535 - p° / 17.535 = 0.139 / (0.139+25) ⇒ 17.535 - p1 = 0.097 ⇒ p1 = 17.44 mm of Hg Hence, the vapour pressure of water is 17.44 mm of Hg.
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7 months ago
```							Vapour pressure of water, p1° = 17.535 mm of HgMass of glucose, w2 = 25 gMass of water, w1 = 450 gWe know that,Molar mass of glucose (C6H12O6),M2 = 6 × 12 + 12 × 1 + 6 × 16 = 180 g mol - 1Molar mass of water, M1 = 18 g mol - 1Then, number of moles of glucose, n1 = 25/180 = 0.139 molAnd, number of moles of water, n2 =450/18 = 25 molNow, we know that,(p1° - p°) / p1° = n1 / n2 + n1⇒ 17.535 - p° / 17.535 = 0.139 / (0.139+25)⇒ 17.535 - p1 = 0.097⇒ p1 = 17.44 mm of HgHence, the vapour pressure of water is 17.44 mm of Hg
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7 months ago
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