SELECT * FROM question_mgmt as q WHERE id=1576 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:

The vapour pressure of water is 12.3 kPa at 300 K. Calculate vapour pressure of 1 molal solution of a non-volatile solute in it.

Answer:

1 molal solution means 1 mol of the solute is present in 1000 g of the solvent (water).

Molar mass of water = 18 g mol - 1

∴ Number of moles present in 1000 g of water = 1000/18

= 55.56 mol

Therefore, mole fraction of the solute in the solution is

x2 =  1 / (1+55.56) = 0.0177.

It is given that,

Vapour pressure of water, p10 = 12.3 kPa

Applying the relation, (P10 -  P1) / P10 = X2 

⇒ (12.3 - p1) / 12.3 =0.0177

⇒ 12.3 - P1 = 0.2177

⇒ p1 = 12.0823

= 12.08 kPa (approximately)

Hence, the vapour pressure of the solution is 12.08 kPa.


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