The vapour pressure of pure water is 23.5 mm Hg. Then, the vapour pressure of an aqueous solution which contains 5 mass percent of urea is (Molar mass of urea is 60).

Question

The vapour pressure of pure water is 23.5 mm Hg. Then, the vapour pressure of an aqueous solution which contains 5 mass percent of urea is (Molar mass of urea is 60). (A) 23 mm Hg (B) 18 mm Hg (C) 31 mm Hg (D) 35 mm Hg

Tardigrade Admin · Accepted Answer

Mass of urea =5 g Mass of water =100-5=95 g Number of moles of urea =(5/60)=0.083 Number of moles of water =(95/18)=5.278 ∴ Total number of moles =5.278+0.083=5.361 Mole fraction of solvent =(5.278/5.361) beginaligned P s &= text Mole fraction of solvent × P °=(5.278/5.361) × 23.5 &=23.14 mm Hg ≈ 23 mm Hg endaligned

Q. The vapour pressure of pure water is 23.5mmHg. Then, the vapour pressure of an aqueous solution which contains 5 mass percent of urea is (Molar mass of urea is 60).

23 mm Hg

18 mm Hg

31 mm Hg

35 mm Hg

Mass of urea =5g Mass of water =100−5=95g Number of moles of urea =605​=0.083 Number of moles of water =1895​=5.278 ∴ Total number of moles =5.278+0.083=5.361 Mole fraction of solvent =5.3615.278​ Ps​​= Mole fraction of solvent ×P∘=5.3615.278​×23.5=23.14mmHg≈23mmHg​

Q. The vapour pressure of pure water is $23.5 mm H g$ . Then, the vapour pressure of an aqueous solution which contains $5$ mass percent of urea is (Molar mass of urea is $60$ ).