XII-NEET : 2023-24 Don’t Solve this DPP

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DPP- 08 : Solution
XIIth Batch Date : 30-03-2023
1. For a dilute solution, Raoult’s law states that 8. Which of the following 0.1M aqueous solution will
have the lowest vapour pressure ?
1) The lowering of vapour pressure is equal to the
mole fraction of the solute 1) Potassium sulphate
2) The relative lowering of vapour pressure is equal 2) Sodium chloride
to the mole fraction of the solute 3) Urea
3) The relative lowering of vapour pressure is 4) Glucose
proportional to the amount of the solute in 9. Which of the following solutions will exhibit
solution highests lowering vapour pressure ?
4) The vapour pressure of the solution is equal to 1) 0.01 m Na2SO4 2) 0.01 m KNO3
the mole fraction of the solvent
3) 0.015 m urea 4) 0.015 m glucose
2. The relative lowering of the vapour pressure is equal
10. Vapour pressure of dilute aqueous solution of
to the ratio between the numbere of:
glucose is 750 mm of mercury at 373K. The mole
1) Solute molecules to the solvent molecules fraction of the solute is
2) Solute molecules to the total molecules in the 1) 1/76 2) 1/7.6
solution
3) 1/38 4) 1/10
3) Solvent molecules to the total molecules in the
11. The vapour pressure of water at 20°C is 17.5 mm
solution
Hg. If 18 g glucose (C6H12O6) is added to 178.2 g of
4) Solvent molecules to the total number of ions of water at 20°C, the vapour pressure of the resulting
the solute solution will be
3. Which one of the following pais of the solution can 1) 16.500 mm Hg 2) 17.325 mm Hg
we expect to at the same RLVP ?
3) 17.675 mm Hg 4) 15.750 mm Hg
1) 0.1 M urea and 0.1 M NaCl
12. An aqueous solution is 1.00 molal in KI. Which
2) 0.1 M urea and 0.2 M MgCl2 change will cause the vapour pressure of the solution
3) 0.1 M NaCl and 0.1 M Na2SO4 to increase ?
4) 0.1 M Ca(NO3)2 and 0.1 M Na2SO4 1) Addition of water
4. The ratio of the value of any colligative property 2) Addition of NaCl
for KCl solution to that for sugar is nearly _____ 3) Addition of Na2SO4
times.
4) Addition of 1.00 molal KI
1) 1 2) 0.5
13. The vapour pressure of a solvent is decreased by 10
3) 2 4) 2.5 mm of Hg when a non-volatile solute was added to
5. Which of the following solutions (in H2O) has the solvent. The mole fraction of the solute in the
highest L.V.P. ? solution is 0.2. What would be the mole fraction of
1) 0.2 m urea 2) 0.1 m K4[Fe(CN)6] solvent if decrease in vapour pressure is 20 mm of
3) 0.2 m K2SO4 4) 0.3 m glucose Hg ?

6. The vapour pressure of equimolal aqueous solution 1) 0.8 2) 0.6

will be highest for : 3) 0.4 4) 0.2
1) C6H5NH3Cl (aniline hydrochloride) 14. Vapour pressure of pure water at 298K is 23.8 mm
Hg. 50 g of urea (NH2CONH2) is dissolved in 850 g
2) Ca(NO3)2
of water. Vapour pressure of water for this solution
3) La(NO3)3 will be
4) C6H12O6(glucose) 1) 21.4 mm 2) 23.4 mm
7. When mercuric iodide is added to the aqueous 3) 25.4 mm 4) 27.4 mm
solution of potassium iodide, the :
15. 18 g of glucose (C6H12O6) is added to 178.2 g of water.
1) Lowering vapour pressure is raised The vapour pressure of water for this aqueous
2) Lowering vapour pressure is lowered solution at 100°C is
3) Lowering vapour pressure does not change 1) 76.00 torr 2) 752.40 Torr
4) Boiling point does not change 3) 759.00 Torr 4) 7.60 Torr
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16. Vapour pressure of CCl4 at 25°C is 143 mm Hg. 0.5 g 25. Which fo the following is the expression of Raoult’s
of a non-volatile solute (mol. wt. 65) is dissolved in law ?
100 cm3 of CCl4. Find the vapour pressure of the (p = vapour pressure of pure solvent)
solution. (Density of CCl4 = 1.58 g/cm3)

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ps = vapour pressure of pure solution
1) 141.93 mm 2) 94.39 mm
3) 199.34 mm 4) 143.99 mm p  ps n ps  p N
1)  2) 
17. Which one of the following is an expression of p nN p Nn
Raoult’s law if PA is the partial pressure of the solvent
in a solution. P°A is the partial pressure of the pure p  ps N ps  p N  n
3)  4) 
solvent and if XA and XB are the mole fraciton of the ps Nn ps N
solute and the solvent respectively?
26. The mas sof non-volatile slute of molar mass 40 g/
1) PA = P°AXA 2) PA = P°AXA(1/XB)
mol that should be dissolved in 114 g of octane to
3) PA = P°AXB 4) P°A = P(XA/XB) lower its vapour pressure by 20% is
18. The vapour pressur eo fwater at room temperature 1) 11.4 g 2) 9.8 g
is 23.8 mm Hg. The vapour pressur eof an aqueous
3) 12.8 g 4) 8 g
solution of sucrose with mole fraction 0.1 is equal
to 27. Lowering of vapour pessur eof an aqueous solution
of a non-volatile, non-electrolyte 1m aqueous
1) 3.9 mm Hg 2) 24.2 Hg
solution at 100°C is
3) 21.42 mm Hg 4) 21.44 mm Hg
1) 14.12 torr 2) 312 torr
19. 34.2 g of canesugar is dissolved in 180 g of water.
3) 13.45 torr 4) 352 torr
The relative lowering of vapour pressure will be
28. The relative lowering of vapour pressure is equal
1) 0.0099 2) 1.1597
to the ratio between the number of
3) 0.840 4) 0.9901
1) solute molecule to the solvent molecules
20. The relative lowering of vapour pressure produdced
2) solute moleucles to the total moleucles in the
by a solution of 18 g of urea in 100 g of water is
solution
approximately
3) solvent molecules to the total molecules in the
1) 0.5 2) 0.05
solution
3) 0.25 4) 0.025
4) solvent molecules to the total number of ions of
21. 3g of urea is dissolved in 45 g of water the relative the solute
lowering of vapour pressure is
29. 60 gm of Urea (Mol. wt 60) was dissolved in 9.9
1) 0.05 2) 0.04 moles of water. If the vapour pressur eof pure water
3) 0.02 4) 0.01 is P0. The vapour pressure of solution is
22. The mass of a non-volatile solute of molar mass 60 1) 0.10 P0 2) 1.10 P0
gmol–1 that should to dissolved in 126 g of water to 3) 0.90 P0 4) 0.99 P0
reduce it vapour pressure to 99% will be
30. The mass of glucose that should be dissolved in 50
1) 4 g 2) 8 g g of water in order to produce the same lowering of
3) 5.6 g 4) 3 g vapour pressure as is produced by dissolving 1g of
23. Vapour pressure of watere at 293 K is 17.535 mm urea in the same quantity of water is
Hg. The vapour pressure of water at 293 K 1) 1g 2) 3g
containing 25 g of glucose dissolved in 450 g of 3) 6g 4) 18g
water is
1) 17.439 mm Hg 2) 17.535 mm Hg
3) 0.097 mm Hg 4) 34.973 mm Hg
24. The vapour pressure lowering caused by the
addition of 100 g of sucrose (molecular mass = 342)
to 1000 g of water, if the vapour pressure of water at
25°C is 23.8 mm Hg is
1) 0.12 mm Hg 2) 0.125 mm Hg
3) 1.15 mm Hg 4) 1.25 mm Hg

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 Answer key with Solutions
DPP-08 Solutions
15. (2)
1. (2)

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2. (2)
3. (4)
4. (3)
5. (3)
6. (4)
7. (1)
8. (1)
9. (1)

10. (1)

11. (2)

16. (1)
17. (3)

18. (3)

19. (1)

12. (1) On addition of water, concentration of the

solution ddecreases, i..e, now less solute particles 20. (2)
are present on the surface or more solvent molecules
are present on the surface. Hence, vapour pressure
increases.
21. (3)

13. (2)

22. (1)

23. (1)
14. (1)

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 27.
26.
25.
24.

(3)
(4)
(1)
(2)

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30.
29.
28.

(2)
(3)
(2)