SELECT * FROM question_mgmt as q WHERE id=1719 AND status=1 SELECT id,question_no,question,chapter FROM question_mgmt as q WHERE courseId=3 AND subId=9 AND chapterId=57 and ex_no='2' AND status=1 ORDER BY CAST(question_no AS UNSIGNED)

Question:

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.

Answer:

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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