Chemistry - Class Twelve

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Solution of NCERT In Text Questions-2.9

Question - 2.9 - Vapour pressure of pure water at 298 K is 23.8 mm Hg. 50 g of urea (NH2 CONH2 ) is dissolved in 850 g of water. Calculate the vapour pressure of water for this solution and its relative lowering.

Solution:

Weight of Urea (WB ) = 50 g

Molar mass (MB ) of Urea (NH2 CONH 2) = 14 + 1 x 2 + 12 +16 +14 + 1 x 2 = 60 g mol – 1

Weight of water (W A) = 850 g

Molar mass of water (M A) = 18 g mol – 1

Vapour pressure of water (P A o) = 23.8 mm Hg

Vapour pressure of water in the given solution P A = ?

Now, number of moles of urea Solutions class 12 chemistry - NCERT In Text Solution49

Solutions class 12 chemistry - NCERT In Text Solution50

Now, number of moles of water Solutions class 12 chemistry - NCERT In Text Solution51

Solutions class 12 chemistry - NCERT In Text Solution52

Now, Mole fraction of urea

Solutions class 12 chemistry - NCERT In Text Solution53

Now, we know that,

Solutions class 12 chemistry - NCERT In Text Solution54

Thus, vapour pressure of water in this solution is 23.40 mm of Hg and its relative lowering is 0.017

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Topics covered in this chapter of Solutions

1. Solutions

2. Types of Solutions

3. Solutions - Concentration

4. Solution of NCERT In Text Questions

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