Test: MCQs (One or More Correct Option): Solutions | JEE Advanced


17 Questions MCQ Test Class 12 Chemistry 35 Years JEE Mains &Advance Past yr Paper | Test: MCQs (One or More Correct Option): Solutions | JEE Advanced


Description
This mock test of Test: MCQs (One or More Correct Option): Solutions | JEE Advanced for JEE helps you for every JEE entrance exam. This contains 17 Multiple Choice Questions for JEE Test: MCQs (One or More Correct Option): Solutions | JEE Advanced (mcq) to study with solutions a complete question bank. The solved questions answers in this Test: MCQs (One or More Correct Option): Solutions | JEE Advanced quiz give you a good mix of easy questions and tough questions. JEE students definitely take this Test: MCQs (One or More Correct Option): Solutions | JEE Advanced exercise for a better result in the exam. You can find other Test: MCQs (One or More Correct Option): Solutions | JEE Advanced extra questions, long questions & short questions for JEE on EduRev as well by searching above.
QUESTION: 1

An azeotropic solution of two liquids has boiling point lowerthan either of them when it      (1981 - 1 Mark)

Solution:

Lower the B. pt., higher will be the V.P. The V.P. of the mixture is greater than either of the two liquids.
[NOTE : : In case of positive deviation from Roult's law the partial vapour pressure of each liquid and total vapour pressure of solution will be greater as compared to initial solution]

QUESTION: 2

For a dilute solution, Raoult’s law states that :(1985 - 1 Mark)

Solution:

[Mathematical statement of Raoult's law]

QUESTION: 3

When mercuric iodide is added to the aqueous solution ofpotassium iodide then (1987 - 1 Mark)

Solution:

Added HgI2 forms a complex with KI in the solution as follows
2KI + HgI2 → K2[HgI4]
As a result, number of particles decreases and so ΔTf increases.
[NOTE : :
Depression in freezing point is a colligative property]

QUESTION: 4

Which of the following 0.1 M aqueous solutions will havethe lowest freezing point? (1989 - 1 Mark)

Solution:

NOTE : The salt producing highest number of ions will have lowest freezing point.

 gives highest number of particles (2 + 1 = 3).
Glucose, being non-electrolyte gives minimum no. of particles and hence minimum ΔTf or maximum F. pt.

QUESTION: 5

The freezing point of equimolal aqueous solutions will behighest for : (1990 - 1 Mark)

Solution:

TIPS/Formulae : The salt that ionises to least extent will have highest freezing point. [ i.e., minimum Δ Tf ]
Glucose, being non electrolyte, gives minimum no. of particles and hence minimum DTf or maximum F. pt

QUESTION: 6

0.2 molal acid HX is 20% ionised in solution. Kf = 1.86 Kmolality–1. The freezing point of the solution is : (1995S)

Solution:

Depression in freezing point, 

Van’t Hoff factor,   where n = no. of ions

[ For 20 % ionisation, α = 20/100= 0.2]
∴ΔTf = 1.2 × 1.86 × 0.2 = 0.45 [ ∵ m = 0.2]
Hence freezing point of solution is 0 – 0.45 = –0.45
[∵ F.P of water = 0.0 C]

QUESTION: 7

The molecular weight of benzoic acid in benzene asdetermined by depression in freezing point methodcorresponds to : (1996 - 1 Mark)

Solution:

Benzoic acid exists as dimer in benzene.

[Normal molecular mass = 122 amu
observed molecular mass = 244 amu, in case of complete association] 

QUESTION: 8

During depression of freezing point in a solution thefollowing are in equililbrium (2003S)

Solution:

NOTE : At the freezing point liquid and solid remain in equilibrium. If a solution of a non-volatile solute is cooled to a temperature below the freezing point of solution, some of liquid solvent will separate as a solid solvent and thus the concentration of solution will
increase.

QUESTION: 9

The elevation in boiling point of a solution of 13.44 g ofCuCl2 in 1 kg of water using the following information will be(Molecular weight of CuCl2 = 134.4 and Kb = 0.52 K molal-1)       (2005S)

Solution:

TIPS/Formulae :

Assuming 100% ionization So, i = 1 + 2 = 3

QUESTION: 10

When 20 g of naphthoic acid (C11H8O2) is dissolved in 50g of benzene (Kf = 1.72 K kg mol–1), a freezing pointdepression of 2K is observed. The Van't Hoff factor (i) is            (2007)

Solution:

Molecular weight of naphthoic acid
C11H8O2 = 172 gmol–1.
The theoretical value of depression in freezing point

Van't Hoff factor,

QUESTION: 11

The Henry’s law constant for the solubility of N2 gas inwater at 298 K is 1.0 × 105 atm. The mole fraction of N2 in airis 0.8. The number of moles of N2 from air dissolved in10 moles of water at 298 K and 5 atm pressure is (2009)

Solution:

QUESTION: 12

Dissolving 120 g of urea (mol. wt. 60) in 1000 g of water gavea solution of density 1.15 g/mL. The molarity of the solutionis (2011)

Solution:

Number of moles of urea = 120/60 = 2

Total mass of solution = 1000 + 120 = 1120 g
Total volume of solution (in L) = Mass/density

Molarity of the solution

 

QUESTION: 13

The freezing point (in °C) of a solution containing 0.1 gof K3[Fe(CN)6] (Mol. wt. 329) in 100 g of water (Kf = 1.86 Kkg mol–1) is (2011)

Solution:

ΔTf = i × Kf × m
Where m = Molality of the solution
(i.e. number of moles of solute per 1000 g of the solvent)

QUESTION: 14

For a dilute solution containing 2.5 g of a non-volatile nonelectrolytesolute in 100 g of water, the elevationin boiling point at 1 atm pressure is 2°C. Assumingconcentration of solute is much lower than the concentrationof solvent, the vapour pressure (mm of Hg) of the solutionis (take Kb = 0.76 K kg mol–1) (2012)

Solution:

From Raoult law relation,

When the concentration of solute is much lower than the concentration of solvent,

           .... (i)

From elevation in boiling point, ΔTb = Kb × m
2 = 0.76 × m
m = 2/0.76              ...(ii)
From (i) and (ii), p = 724 mm

*Multiple options can be correct
QUESTION: 15

In the depression of freezing point experiment, it is foundthat the (1999 - 3 Marks)

Solution:

(a, d) The freezing point of a solvent depresses as a nonvolatile solute is added to a solvent. According to Raoult's law, when a non-volatile solute is added to a solvent the vapour pressure of the solvent decreases.
At the freezing point it will be only the solvent
molecules which will solidify.

*Multiple options can be correct
QUESTION: 16

Benzene and naphthalene form an ideal solution at roomtemperature. For this process, the true statement(s) is(are)                 (JEE Adv. 2013)

Solution:

*Multiple options can be correct
QUESTION: 17

Mixture(s) showing positive deviation from Raoult’s law at35 °C is (are) (JEE Adv. 2016)

Solution:

(A) H-bonding of methanol breaks when CCl4 is added so bonds become weaker, resulting positive deviation.
(B) Mixing of polar and non-polar liquids will produce a solution of weaker interaction, resulting positive deviation
(C) Ideal solution
(D) –ve deviation because stronger H-bond is formed.