Vapour pressure of water at 293 Kis 17.535 mm Hg. Calculate the vapour pressure of water at 293 Kwhen 25 g of glucose is dissolved in 450 g of water.
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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very good
Did\'nt understand the last second step how does 0.097 comes? Please answer....
Bole to jhakkas
thnx
Irrelevant answer .. Waste of time... Not at all satisfied...