Chemistry Class Twelve

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Solutions

Solution of NCERT In Text Questions (Part - 3)

Question - 2.8 The vapour pressure of pure liquids A and B are 450 and 700 mm Hg respectively, at 350 K . Find out the composition of the liquid mixture if total vapour pressure is 600 mm Hg. Also find the composition of the vapour phase.

Solution:

Given,

Solutions class 12 chemistry - NCERT In Text Solution44

For liquid A and B,

From Rault’s Law,

Solutions class 12 chemistry - NCERT In Text Solution45
Solutions class 12 chemistry - NCERT In Text Solution46

Now, In vapour phase,

Mole fraction of liquid A

Solutions class 12 chemistry - NCERT In Text Solution47

And mole fraction of liquid B

Solutions class 12 chemistry - NCERT In Text Solution48

Thus, composition in liquid phase, is 0.4 and 0.6

And that in vapour phase, is 0.3 and 0.7

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

Question - 2.10 Boiling point of water at 750 mm Hg is 99.63°C. How much sucrose is to be added to 500 g of water such that it boils at 100°C.

Solution:

Given,

Elevation of boiling point

Solutions class 12 chemistry - NCERT In Text Solution55

Mass of water (W A) = 500 g

Molar mass (M B) of sucrose (C 12H22 O11 ) = 12 x 12 + 1 x 22 + 16 x 11 = 342 g mol – 1

Molal constant for water (Kb ) = 0.52 K kg mol – 1

Therefore, W B = ?

We know that,

Solutions class 12 chemistry - NCERT In Text Solution56

Thus, 121.67 g of sucrose to be added.

Question – 2.11 - Calculate the mass of ascorbic acid (Vitamin C, C6 H8 O 6) to be dissolved in 75 g of acetic acid to lower its melting point by 1.5°C. K f = 3.9 K kg mol-1.

Solution:

Given,

Lowering of melting point

Solutions class 12 chemistry - NCERT In Text Solution57

Molar mass of ascorbic acid (M B) = 176 g mol – 1

Thus, mass of ascorbic (WB ) = ?

We know that,

Solutions class 12 chemistry - NCERT In Text Solution58

Thus, required mass of ascorbic acid = 5.077 g

Question - 2.12 Calculate the osmotic pressure in pascals exerted by a solution prepared by dissolving 1.0 g of polymer of molar mass 185,000 in 450 mL of water at 37°C.

Solution:

Given, Mass (W B) of polymer = 1 g

Molar mass (MB ) of polymer = 185000

Volume of water (V) = 450 mL = 0.45 L

Temperature (T) = 37 0C = 37+273 = 310 K

Osmotic pressure = ?

We know, Solutions class 12 chemistry - NCERT In Text Solution59

Number of moles of the polymer (n) Solutions class 12 chemistry - NCERT In Text Solution60

We know that Osmotic pressure Solutions class 12 chemistry - NCERT In Text Solution61

Solutions class 12 chemistry - NCERT In Text Solution62

Thus, Osmotic pressure = 30.95 Pa

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Solutions - Class 12 chemisty - NCERT In Text Solution (Part 1)
Solutions - Class 12 chemisty - NCERT In Text Solution (Part 2)


 
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