iCON Education - Solutions

2018 JEE Mains MCQ

iCON Education HYD, 79930 92826, 73309 72826 JEE Main 2018 (Online) 16th April Morning Slot

The mass of a non-volatile, non-electrolyte solute (molar mass = 50 g mol^-1 ) needed to be dissolved in 114 g octane to reduce its vapour pressure to 75%, is :

A.

37.5 g

B.

75 g

C.

150 g

D.

50 g

2018 JEE Mains MCQ

iCON Education HYD, 79930 92826, 73309 72826 JEE Main 2018 (Offline)

For 1 molal aqueous solution of the following compounds, which one will show the highest freezing point?

A.

[Co(H₂O)₃ Cl₃].3H₂O

B.

[Co(H₂O)₆] Cl₃

C.

[Co(H₂O)₅ Cl] Cl₂.H₂O

D.

[Co(H₂O)₄ Cl₂] Cl.2H₂O

2018 JEE Mains MCQ

iCON Education HYD, 79930 92826, 73309 72826 JEE Main 2018 (Online) 15th April Evening Slot

Two 5 molal solutions are prepared by dissolving a non-electrolyte non-volatile solute separately in the solvents X and Y. The molecular weights of the solvents are M_x and M_y, respectively where M_x = ${3 \over 4}$ My. The relative lowering of vapor pressure of the solution in X is ''m'' times that of the solution in Y. Given that the number of moles of solute is very small in comparison to that of the solvent, the value of ''m'' is :

A.

${4 \over 3}$

B.

${3 \over 4}$

C.

${1 \over 2}$

D.

${1 \over 4}$

2017 JEE Mains MCQ

iCON Education HYD, 79930 92826, 73309 72826 JEE Main 2017 (Online) 9th April Morning Slot

A solution is prepared by mixing 8.5 g of CH₂Cl₂ and 11.95 g of CHCl₃ . If vapour pressure of CH₂Cl₂ and CHCl₃ at 298 K are 415 and 200 mmHg respectively, the mole fraction of CHCl₃ in vapour form is : (Molar mass of Cl = 35.5 g mol⁻¹)

A.

0.162

B.

0.675

C.

0.325

D.

0.486

2017 JEE Mains MCQ

iCON Education HYD, 79930 92826, 73309 72826 JEE Main 2017 (Online) 8th April Morning Slot

5 g of Na₂SO₄ was dissolved in x g of H₂O. The change in freezing point was found to be 3.82^oC. If Na₂SO₄ is 81.5% ionised, the value of x

(K_f for water=1.86^oC kg mol⁻¹) is approximately :

(molar mass of S = 32 g mol⁻¹ and that of Na = 23 g mol⁻¹)

A.

15 g

B.

25 g

C.

45 g

D.

65 g

Correct Answer: C

Explanation:

	Na₂SO₄	$\rightleftharpoons$	2Na⁺	+	SO₄^-2
Initial	1 mol		0		0
After dissociation	1 - x		2x		x

$\therefore\,\,\,$ Total no of Moles after dissociation = 1 + 2x

Na₂SO₄ is ionised 81.5% means x = 0.815

Von't Hoff factor (i) = ${{Moles\,\,after\,\,dissociation} \over {Initial\,\,no.\,\,of\,\,moles}}$

= ${{1 + 2x} \over 1}$

= 1 + 2 $ \times $ 0.815

= 2.63

$\therefore\,\,\,$ $\Delta $T_f = ${{1000 \times {K_f} \times {w_2} \times i} \over {{M_S} \times {w_1}}}$

$ \Rightarrow $$\,\,\,$ 3.82 = ${{1000 \times 1.86 \times 2.63 \times 5} \over {142 \times x}}$

$ \Rightarrow $$\,\,\,$ x = 45 gm

2017 JEE Mains MCQ

iCON Education HYD, 79930 92826, 73309 72826 JEE Main 2017 (Offline)

The freezing point of benzene decreases by 0.45⁰C when 0.2 g of acetic acid is added to 20g of benzene. If acetic acid associates to form a dimer in benzene, percentage association of acetic acid in benzene will be: (K_f for benzene = 5.12 K kg mol^–1)

A.

80.4 %

B.

74.6 %

C.

94.6 %

D.

64.6 %

2016 JEE Mains MCQ

iCON Education HYD, 79930 92826, 73309 72826 JEE Main 2016 (Online) 10th April Morning Slot

An aqueous solution of a salt MX₂ at certain temperature has a van’t Hoff factor of 2. The degree of dissociation for this solution of the salt is :

A.

0.33

B.

0.50

C.

0.67

D.

0.80

2016 JEE Mains MCQ

iCON Education HYD, 79930 92826, 73309 72826 JEE Main 2016 (Online) 9th April Morning Slot

The solubility of N₂ in water at 300 K and 500 torr partial pressure is 0.01 g L⁻¹. The solubility (in g L⁻¹) at 750 torr partial pressure is :

A.

0.0075

B.

0.015

C.

0.02

D.

0.005

2016 JEE Mains MCQ

iCON Education HYD, 79930 92826, 73309 72826 JEE Main 2016 (Offline)

18 g glucose (C₆H₁₂O₆) is added to 178.2 g water. The vapor pressure of water (in torr) for this aqueous solution is :

A.

76.0

B.

752.4

C.

759.0

D.

7.6

2015 JEE Mains MCQ

iCON Education HYD, 79930 92826, 73309 72826 JEE Main 2015 (Offline)

The vapour pressure of acetone at 20^oC is 185 torr. When 1.2 g of a non-volatile substance was dissolved in 100 g of acetone at 20^oC, its vapour pressure was 183 torr. The molar mass (g mol^-1) of the substance is:

A.

64

B.

128

C.

488

D.

32

2014 JEE Mains MCQ

iCON Education HYD, 79930 92826, 73309 72826 JEE Main 2014 (Offline)

Consider separate solutions of 0.500 M C₂H₅OH(aq), 0.100 M Mg₃(PO₄)₂(aq), 0.250 M KBr(aq) and 0.125 M Na₃PO₄(aq) at 25^oC. Which statement is true about these solutions, assuming all salts to be strong electrolytes?

A.

0.125 M Na₃PO₄(aq) has the highest osmotic pressure.

B.

0.500 M C₂H₅OH(aq) has the highest osmotic pressure

C.

They all have the same osmotic pressure

D.

0.100 M Mg₃(PO₄)₂(aq) has the highest osmotic pressure.

2012 JEE Mains MCQ

iCON Education HYD, 79930 92826, 73309 72826 AIEEE 2012

K_f for water is 1.86K kg mol^–1. If your automobile radiator holds 1.0 kg of water, how many grams of ethylene glycol (C₂H₆O₂) must you add to get the freezing point of the solution lowered to –2.8^oC ?

A.

72 g

B.

93 g

C.

39 g

D.

27 g

2011 JEE Mains MCQ

iCON Education HYD, 79930 92826, 73309 72826 AIEEE 2011

The degree of dissociation ($\alpha$ ) of a weak electrolyte, A_xB_y is related to van’t Hoff factor (i) by the expression :

A.

$\alpha = {{i - 1} \over {x + y + 1}}$

B.

$\alpha = {{x + y - 1} \over {i - 1}}$

C.

$\alpha = {{x + y + 1} \over {i - 1}}$

D.

$\alpha = {{i - 1} \over {(x + y - 1)}}$

2011 JEE Mains MCQ

iCON Education HYD, 79930 92826, 73309 72826 AIEEE 2011

Ethylene glycol is used as an antifreeze in a cold climate. Mass of ethylene glycol which should be added to 4 kg of water to prevent it from freezing at −6^oC will be : [K_f for water = 1.86 K kg mol⁻¹ , and molar mass of ethylene glycol = 62 g mol⁻¹ )

A.

204.30 g

B.

400.00 g

C.

304.60 g

D.

804.32 g

2010 JEE Mains MCQ

iCON Education HYD, 79930 92826, 73309 72826 AIEEE 2010

On mixing, heptane and octane form an ideal solution. At 373 K, the vapour pressures of the two liquid components (heptane and octane) are 105 kPa and 45 kPa respectively. Vapour pressure of the solution obtained by mixing 25.0g of heptane and 35 g of octane will be (molar mass of heptane = 100 g mol^–1 and of octane = 114 g mol^–1)

A.

72.0 kPa

B.

36.1 kPa

C.

96.2 kPa

D.

144.5 kPa

2010 JEE Mains MCQ

iCON Education HYD, 79930 92826, 73309 72826 AIEEE 2010

If sodium sulphate is considered to be completely dissociated into cations and anions in aqueous solution, the change in freezing point of water (∆T_f), when 0.01 mol of sodium sulphate is dissolved in 1 kg of water, is (K_f = 1.86 K kg mol^–1)

A.

0.0372 K

B.

0.0558 K

C.

0.0744 K

D.

0.0186 K

2009 JEE Mains MCQ

iCON Education HYD, 79930 92826, 73309 72826 AIEEE 2009

Two liquids X and Y form an ideal solution. At 300K, vapour pressure of the solution containing 1 mol of X and 3 mol of Y is 550 mm Hg. At the same temperature, if 1 mol of Y is further added to this solution, vapour pressure of the solution increases by 10 mm Hg. Vapour pressure (in mm Hg) of X and Y in their pure states will be, respectively :

A.

200 and 300

B.

300 and 400

C.

400 and 600

D.

500 and 600

2009 JEE Mains MCQ

iCON Education HYD, 79930 92826, 73309 72826 AIEEE 2009

A binary liquid solution is prepared by mixing n-heptane and ethanol. Which one of the following statements is correct regarding the behaviour of the solution ?

A.

The solution formed is an ideal solution

B.

The solution is non-ideal, showing +ve deviation from Raoult’s law.

C.

The solution is non-ideal, showing –ve deviation from Raoult’s law.

D.

n-heptane shows +ve deviation while ethanol shows –ve deviation from Raoult’s law.

2008 JEE Mains MCQ

iCON Education HYD, 79930 92826, 73309 72826 AIEEE 2008

The vapour pressure of water at 20^oC is 17.5 mm Hg. If 18 g of glucose (C₆H₁₂O₆) is added to 178.2 g of water at 20^oC, the vapour pressure of the resulting solution will be

A.

17.675 mm Hg

B.

15.750 mm Hg

C.

16.500 mm Hg

D.

17.325 mm Hg

2008 JEE Mains MCQ

iCON Education HYD, 79930 92826, 73309 72826 AIEEE 2008

At 80^oC, the vapour pressure of pure liquid ‘A’ is 520 mm Hg and that of pure liquid ‘B’ is 1000 mm Hg. If a mixture solution of ‘A’ and ‘B’ boils at 80^oC and 1 atm pressure, the amount of ‘A’ in the mixture is (1 atm = 760 mm Hg)

A.

52 mol percent

B.

34 mol percent

C.

48 mol percent

D.

50 mol percent

2007 JEE Mains MCQ

iCON Education HYD, 79930 92826, 73309 72826 AIEEE 2007

A mixture of ethyl alcohol and propyl alcohol has a vapour pressure of 290 mm at 300 K. The vapour pressure of propyl alcohol is 200 mm. If the mole fraction of ethyl alcohol is 0.6, its vapour pressure (in mm) at the same temperature will be

A.

360

B.

350

C.

300

D.

700

2007 JEE Mains MCQ

iCON Education HYD, 79930 92826, 73309 72826 AIEEE 2007

A 5.25 % solution of a substance is isotonic with a 1.5% solution of urea (molar mass = 60 g mol⁻¹) in the same solvent. If the densities of both the solutions are assumed to be equal to 1.0 g cm⁻³, molar mass of the substance will be

A.

90.0 g mol⁻¹

B.

115.0 g mol⁻¹

C.

105.0 g mol⁻¹

D.

210.0 g mol⁻¹

2006 JEE Mains MCQ

iCON Education HYD, 79930 92826, 73309 72826 AIEEE 2006

18 g of glucose (C₆H₁₂O₆) is added to 178.2 g of water. The vapour pressure of water for this aqueous solution at 100^oC is

A.

759.00 Torr

B.

7.60 Torr

C.

76.00 Torr

D.

752.40 Torr

2006 JEE Mains MCQ

iCON Education HYD, 79930 92826, 73309 72826 AIEEE 2006

Among the following mixtures, dipole-dipole as the major interaction, is present in

A.

benzene and ethanol

B.

acetonitrile and acetone

C.

KCl and water

D.

benzene and carbon tetrachloride

2005 JEE Mains MCQ

iCON Education HYD, 79930 92826, 73309 72826 AIEEE 2005

If $\alpha$ is the degree of dissociation of Na₂SO₄, the vant Hoff’s factor (i) used for calculating the molecular mass is :

A.

1 + $\alpha$

B.

1 + $2\alpha$

C.

1 - $\alpha$

D.

1 - $2\alpha$

2005 JEE Mains MCQ

iCON Education HYD, 79930 92826, 73309 72826 AIEEE 2005

Benzene and toluene form nearly ideal solutions. At 20 ^oC, the vapour pressure of benzene is 75 torr and that of toluene is 22 torr. The partial vapour pressure of benzene at 20 ^oC for a solution containing 78 g of benzene and 46 g of toluene in torr is

A.

50

B.

25

C.

53.5

D.

37.5

2005 JEE Mains MCQ

iCON Education HYD, 79930 92826, 73309 72826 AIEEE 2005

Equimolar solutions in the same solvent have

A.

Same boiling point but different freezing point

B.

Same freezing point but different boiling point

C.

Same boiling and same freezing points

D.

Different boiling and different freezing points

2004 JEE Mains MCQ

iCON Education HYD, 79930 92826, 73309 72826 AIEEE 2004

Which one of the following aqueous solutions will exhibit highest boiling point?

A.

0.01 M Na₂SO₄

B.

0.015 M glucose

C.

0.015 M urea

D.

0.01 M KNO₃

2004 JEE Mains MCQ

iCON Education HYD, 79930 92826, 73309 72826 AIEEE 2004

For which of the following parameters the structural isomers C₂H₅OH and CH₃OCH₃ would be expected to have the same values? (Assume ideal behaviour)

A.

Heat of vaporization

B.

Gaseous densities at the same temperature and pressure

C.

Boiling points

D.

Vapour pressure at the same temperature

2004 JEE Mains MCQ

iCON Education HYD, 79930 92826, 73309 72826 AIEEE 2004

Which one of the following statements is false?

A.

Raoult’s law states that the vapour pressure of a components over a solution is proportional to its mole fraction

B.

Two sucrose solutions of same molality prepared in different solvents will have the same freezing point depression

C.

The correct order of osmotic pressure for 0.01 M aqueous solution of each compound is BaCl₂ > KCl > CH₃COOH > sucrose

D.

The osmotic pressure ($\pi$) = MRT, where M is the molarity of the solution

2004 JEE Mains MCQ

iCON Education HYD, 79930 92826, 73309 72826 AIEEE 2004

Which of the following liquid pairs shows a positive deviation from Raoult’s law?

A.

Water – hydrochloric acid

B.

Acetone – chloroform

C.

Water – nitric acid

D.

Benzene – methanol

2003 JEE Mains MCQ

iCON Education HYD, 79930 92826, 73309 72826 AIEEE 2003

In a 0.2 molal aqueous solution of a weak acid HX the degree of ionization is 0.3. Taking k_f for water as 1.85, the freezing point of the solution will be nearest to

A.

-0.360^oC

B.

-0.260^oC

C.

+0.480^oC

D.

-0.480^oC

2003 JEE Mains MCQ

iCON Education HYD, 79930 92826, 73309 72826 AIEEE 2003

A pressure cooker reduces cooking time for food because

A.

boiling point of water involved in cooking is increased

B.

the higher pressure inside the cooker crushes the food material

C.

cooking involves chemical changes helped by a rise in temperature

D.

heat is more evenly distributed in the cooking space

2003 JEE Mains MCQ

iCON Education HYD, 79930 92826, 73309 72826 AIEEE 2003

If liquids A and B form an ideal solution

A.

the entropy of mixing is zero

B.

the free energy of mixing is zero

C.

the free energy as well as the entropy of mixing are each zero

D.

the enthalpy of mixing is zero

2002 JEE Mains MCQ

iCON Education HYD, 79930 92826, 73309 72826 AIEEE 2002

In a mixture of A and B, components show negative deviation when :

A.

$\Delta V_{mix}$ > 0, $\Delta S_{mix}$ > 0

B.

$\Delta V_{mix}$ = 0, $\Delta S_{mix}$ > 0

C.

A - B interaction is weaker than A - A and B - B interaction

D.

A - B interaction is stronger than A - A and B - B interaction

Correct Answer: D

Explanation:

In the context of solutions and mixtures, a negative deviation from Raoult's law occurs when the interaction between different molecules (A-B interaction) is stronger than the interaction between similar molecules (A-A and B-B interactions). This is because the molecules prefer to interact with each other rather than themselves, leading to a decrease in the overall vapor pressure compared to what would be expected from Raoult's law.

So, the correct option is :

Option D : A - B interaction is stronger than A - A and B - B interaction

Option A suggests that the change in volume ($\Delta V_{mix}$) and change in entropy ($\Delta S_{mix}$) of the mixture are both greater than zero. This doesn't necessarily indicate a negative deviation from Raoult's law. It's possible for these conditions to be true in both positive and negative deviations or even in ideal solutions. Therefore, this option isn't specifically associated with negative deviations.

Option B suggests that there's no change in the volume of the mixture ($\Delta V_{mix} = 0$) and the change in entropy ($\Delta S_{mix}$) is greater than zero. Again, this doesn't specifically indicate a negative deviation from Raoult's law. The volume change upon mixing can be zero, positive, or negative depending on the nature of the interactions among the molecules.

Raoult's law deviations are primarily dictated by the relative strength of the intermolecular forces among the different components in the mixture. Thus, the change in volume and entropy, while they can be related, do not directly determine whether a solution will show a positive or negative deviation.

2002 JEE Mains MCQ

iCON Education HYD, 79930 92826, 73309 72826 AIEEE 2002

Freezing point of an aqueous solution is (-0.186)^oC. Elevation of boiling point of the same solution is K_b = 0.512 ^oC, K_f = 1.86 ^oC, find the increase in boiling point.

A.

0.186 ^oC

B.

0.0512 ^oC

C.

0.092 ^oC

D.

0.2732 ^oC

Chemistry Chapters

Solutions

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