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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.

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.

Grade:12

3 Answers

Saurabh Koranglekar
askIITians Faculty 10335 Points
4 years ago
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Arun
25750 Points
4 years 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.
Vikas TU
14149 Points
3 years 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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