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Grade: 12

                        

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.

7 months ago

Answers : (3)

Saurabh Koranglekar
askIITians Faculty
10233 Points
							576-431_IMG_20200415_222750.jpg
						
7 months ago
Arun
25089 Points
							
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.
7 months ago
Vikas TU
13786 Points
							
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
7 months ago
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