JEE Exam  >  JEE Tests  >  Revisal Problems For JEE Advanced (Liquid Solutions) - JEE MCQ

Revisal Problems For JEE Advanced (Liquid Solutions) - JEE MCQ


Test Description

30 Questions MCQ Test - Revisal Problems For JEE Advanced (Liquid Solutions)

Revisal Problems For JEE Advanced (Liquid Solutions) for JEE 2024 is part of JEE preparation. The Revisal Problems For JEE Advanced (Liquid Solutions) questions and answers have been prepared according to the JEE exam syllabus.The Revisal Problems For JEE Advanced (Liquid Solutions) MCQs are made for JEE 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Revisal Problems For JEE Advanced (Liquid Solutions) below.
Solutions of Revisal Problems For JEE Advanced (Liquid Solutions) questions in English are available as part of our course for JEE & Revisal Problems For JEE Advanced (Liquid Solutions) solutions in Hindi for JEE course. Download more important topics, notes, lectures and mock test series for JEE Exam by signing up for free. Attempt Revisal Problems For JEE Advanced (Liquid Solutions) | 32 questions in 64 minutes | Mock test for JEE preparation | Free important questions MCQ to study for JEE Exam | Download free PDF with solutions
Revisal Problems For JEE Advanced (Liquid Solutions) - Question 1

Only One Option Correct Type
This section contains 17 multiple choice questions. Each question has four choices (a), (b), (c) and (d), out of which ONLY ONE is correct

 

Q.

Mole fraction of non-electrolyte (A) in aqueous solution is 0.07. One molal urea solution freezes at - 1.86°. Thus, depression in freezing point of the non-electrolyte A is

Detailed Solution for Revisal Problems For JEE Advanced (Liquid Solutions) - Question 1

Revisal Problems For JEE Advanced (Liquid Solutions) - Question 2

A complex of potassium, iron and cyanide is 100% ionised at 1 m (molal). If the elevation in boiling point is 2.08°, then complex is [Kb (H20 ) = 0.52° mol-1 kg]

Detailed Solution for Revisal Problems For JEE Advanced (Liquid Solutions) - Question 2

where, y = number of ions of the complex

x = degree of ionisation = 1

 ∴       y = 4

Four ions are formed from

1 Crore+ students have signed up on EduRev. Have you? Download the App
Revisal Problems For JEE Advanced (Liquid Solutions) - Question 3

Select the correct statement(s).

Detailed Solution for Revisal Problems For JEE Advanced (Liquid Solutions) - Question 3

Thus, (a) is incorrect

(b) Based on (a),   Thus, (b) is correct.

(c) Elevation in boiling point (ΔTb) is a colligative property. Boiling point is not a colligative property. Thus, (c) is incorrect.

Revisal Problems For JEE Advanced (Liquid Solutions) - Question 4

What mass of sucrose, C12H22011, should you dissolve in 100.0 g water to obtain a solution in which mole fraction of C12H22011 is 0.124?

Detailed Solution for Revisal Problems For JEE Advanced (Liquid Solutions) - Question 4

Revisal Problems For JEE Advanced (Liquid Solutions) - Question 5

The freezing point (in °C) of a solution containing 0.1 g of K3[Fe(CN)6] in 100 g water in which it is 80% ionised is [Kf (H20) = 1.86° mol-1 kg]

Detailed Solution for Revisal Problems For JEE Advanced (Liquid Solutions) - Question 5

Revisal Problems For JEE Advanced (Liquid Solutions) - Question 6

Which represents correct difference?

Detailed Solution for Revisal Problems For JEE Advanced (Liquid Solutions) - Question 6

. (a) (I) Entropy change between solution and vapour is smaller than the entropy change between pure solvent and vapour, thus correct.

(II) Heats of vaporisation for a pure solvent and for a solution are similar because similar intermolecular forces between solvent molecules must be overcome in both cases, thus correct.

(III) When a non-volatile solute is added to a solvent, vapour pressure falls hence, boiling point is elevated, thus correct

Revisal Problems For JEE Advanced (Liquid Solutions) - Question 7

60 g of urea is dissolved in 1100 g of solution in water. To keep ΔTf/Kf as 1 mol kg-1, water separated in the form of ice is 

Detailed Solution for Revisal Problems For JEE Advanced (Liquid Solutions) - Question 7

(a)

Solute (urea) = 60 g

Solution (urea + water) = 1100 g

∴ Solvent (water) = 1100- 60 = 1040 g

Thus, water separated in the form of ice = 1040 - 1000 = 40g

Revisal Problems For JEE Advanced (Liquid Solutions) - Question 8

A solution contains 0.1 mole of acetamide in 1 L of glacial acetic acid. When the solution is cooled, the first crystal that appeared at the freezing point contains the molecule(s) of

Detailed Solution for Revisal Problems For JEE Advanced (Liquid Solutions) - Question 8

(b) When a solute is added to a solvent, freezing point is lowered. Thus, freezing point of solution is less than that of the solvent. Let freezing point of solvent = T1 and freezing point of solution = T2.,

T2 <T1

On lowering temperature, first is attained hence, acetic acid is converted into solid (crystal).

Revisal Problems For JEE Advanced (Liquid Solutions) - Question 9

Two moles of the complex (Cu(NH3)3Ci]Cl are dissolved in 3 moles of H20 . Relative decrease in vapour pressure was found to be 0.50. If this solution is treated with aq. AgN03 solution, then there is formation of

Detailed Solution for Revisal Problems For JEE Advanced (Liquid Solutions) - Question 9

Cl- from 2 moles complex = 2 x 0.5 = 1.0

   ∴ AgCI formed = 1 mol

Revisal Problems For JEE Advanced (Liquid Solutions) - Question 10

Total volume of a mixture of 46 g ethanol (C2H5OH) and 72 g water (H20 ) based on the given figure is

Detailed Solution for Revisal Problems For JEE Advanced (Liquid Solutions) - Question 10

From the graph, partial molar volume of ethanol when

(X = 0.2) = 55cmmol-1 

and that of water when ( x = 0.8) = 14cm3mol-1

Revisal Problems For JEE Advanced (Liquid Solutions) - Question 11

pH of 0.1 M dibasic acid is found to be 3. Hence, its osmotic pressure at temperature T K is

Detailed Solution for Revisal Problems For JEE Advanced (Liquid Solutions) - Question 11

Revisal Problems For JEE Advanced (Liquid Solutions) - Question 12

Water is purified by ion-exchange resin (RH2) in which Ca2+ in hard water is replaced by H+

RH2 + Ca2+ ----- >RCa + 2H+

Water coming out of ion-exchange has pH2. Hence, hardness in ppm of Ca2+ ion is 

Detailed Solution for Revisal Problems For JEE Advanced (Liquid Solutions) - Question 12

Revisal Problems For JEE Advanced (Liquid Solutions) - Question 13

A sample of drinking water is contaminated with CHCI3 to the level of 15 ppm by mass. Thus, mass percentage and molality, respectively are

Detailed Solution for Revisal Problems For JEE Advanced (Liquid Solutions) - Question 13

106g of CHCI3 solution has = 15 g of CHCI3 (106 - 15)g H2O = 106g H20 has = 15 g of CHCI3

103 g of H2O has = 15 x 10-3g CHCI3 = 

∴ Molality = 1.255 x 10-4 mol kg-1

Revisal Problems For JEE Advanced (Liquid Solutions) - Question 14

Based on the given dia gram , which of the fo llow ing statem ents regarding the solu tions of two miscible volatile liquids A and B are correct?

I. Plots AC and BD show that Raoult’s law is obeyed for the solution in which A is solute and B is solvent and as well as for that in which B is solute and A is solvent.

II. Plots CD shows that Dalton’s law of partial pressure is observed for the binary solution of components A and 6.

III. In pure state, vapour pressures of volatile liquid differ by 50 units.

IV. Vapour pressure of a mixture containing 40 mole % of B is about 105 units.

Select the correct answer using the codes given below.

Detailed Solution for Revisal Problems For JEE Advanced (Liquid Solutions) - Question 14

BC = vapour pressure of pure A = 85 units

As A is added into 6, vapour pressure of A decreases and that of 6 increases.

Thus, AC shows Raoult's law for a solution of S in A (solvent) and SO shows Raoult's law for a solution of A in B (solvent).

Thus, (I) is correct.

Total vapour pressure at any composition is given by line CD, thus (II) is correct.

it is same as given by graph.
Thus, (IV) is also correct.

Revisal Problems For JEE Advanced (Liquid Solutions) - Question 15

25 mL of an aqueous solu tion of KCI was fo und to require 20 mL of 1.0 M AgN03 solution when titrated using K2Cr04 as an indicator. Thus, d epression in freezing point of KCI solution with 100% ionisation will be

[Kf (H20 ) = 2.0° mol-1 kg, molarity = molality] 

Detailed Solution for Revisal Problems For JEE Advanced (Liquid Solutions) - Question 15

(b)

Ag + + Cl- → AgCI

When titration is complete,  gives chocolate coloured precipitate with Ag+.

2Ag+ + → Ag2Cr04

This end point is detected.

Revisal Problems For JEE Advanced (Liquid Solutions) - Question 16

A 0.001 molal solution of [Pt(NH3)4CI4] in water had a depression in freezing point of 0.0054°. Kf (H20 ) = 1.86° mol-1 kg (assume 100% ionisation). Thus, com plex is

Detailed Solution for Revisal Problems For JEE Advanced (Liquid Solutions) - Question 16

(c)

Let total number of ions due to ionisa tion of the complex = y

NH3 is always a ligand, hence Cl can be ligand or ionisable ions.

Revisal Problems For JEE Advanced (Liquid Solutions) - Question 17

The complete combustion of 2.40 g of a compound of carbon, hydrogen and oxygen yielded 5.46 g of C02 and 2.23 g of H20 . When 8.69 g o f the compound was dissolved in 281 g of water, the freezing point depression was found to be 0.97°C. Thus, molecular formula of the organic compound is [Kf (H20 ) = 1.86° mol-1 kg)]

Detailed Solution for Revisal Problems For JEE Advanced (Liquid Solutions) - Question 17

Oxygen (by difference) = 100 -(62.05 + 10.32)= 27.63%

*Multiple options can be correct
Revisal Problems For JEE Advanced (Liquid Solutions) - Question 18

One or More than One Options Correct Type
This section contains 4 multiple type questions. Each question has 4 choices (a), (b), (c) and (d), out of which ONE or MORE THAN ONE are correct.

Q.

You have been provided with 10.00% by mass solution of ethanol in water.

At 15°C, density = 0.9831 g mL-1 

At 25°C, density = 0.9804 g mL-1 

Select the correct statement(s).

Detailed Solution for Revisal Problems For JEE Advanced (Liquid Solutions) - Question 18

(a, c)

10% ethanol by mass means

100 g (ethanol + water) solution has = 10 g ethanol

90 g H2O has ethanol = 10 g ethanol

Molality does not involve density, hence same at 15°C and 25°C.
Thus, (c) is correct.

*Multiple options can be correct
Revisal Problems For JEE Advanced (Liquid Solutions) - Question 19

One molal aqueous glucose solution has density 1 .180 gm L-1. Also, Kf / Kb = 3 . Thus,

Detailed Solution for Revisal Problems For JEE Advanced (Liquid Solutions) - Question 19

Glucose solution is = 1 mol kg-1 water

Solute (glucose) = 180g

Solvent (water) = 1000 g

*Multiple options can be correct
Revisal Problems For JEE Advanced (Liquid Solutions) - Question 20

For a mixture of binary c om ponents A and 8 with vapo ur pressures and in pure state, total vapour pressure is

where, XB is mole fraction of component B in the mixture. From this equation, it is concluded that

Detailed Solution for Revisal Problems For JEE Advanced (Liquid Solutions) - Question 20

Thus, vapour pressure of 6 increase as Xs increases and decrease as %A decreases.
Line joining and  represent total vapour pressure.
Thus, (b) and (c) are also correct.
Note Vapour pressure of A  is less than vapour pressure of B . Hence, A is less volatile than B. (d) pt0(a| can also be expressed as

*Multiple options can be correct
Revisal Problems For JEE Advanced (Liquid Solutions) - Question 21

If liquids A and B form an ideal solution, then

Detailed Solution for Revisal Problems For JEE Advanced (Liquid Solutions) - Question 21

(a, c)

Two liquids (A, B) form an ideal solution, when force of attraction between A and 6 (in mixture) is same as between A and A and between 6 and 6. In such case,  Δ Hmix= 0.
Since, disorder increases,  Δ Smix. Also, process is spontaneous at a given temperature,

ΔG = ΔH - T ΔS = 0 — T ΔS
ΔGmix < 0

Revisal Problems For JEE Advanced (Liquid Solutions) - Question 22

Comprehension Type
This section contains 3 paragraphs, each describing theory, experiments, data, etc. Two questions related to the paragraph have been given.six question related to paragraphs have been given Each question has only one correct answer out of the given 4 options (a), (b), (c) and (d)

Passage 1

Water boils at 100°C at 1 atm pressure. Latent heat of vaporisation (ΔH ) = 10 kcal mol-1.

Q.

What is the vapour pressure at 40°C?

Detailed Solution for Revisal Problems For JEE Advanced (Liquid Solutions) - Question 22

Variation of vapour pressure with temperature is given Clapeyron-Clausius equation.

p1 = 0.07656 atm

p1 = 0.07656 x 760 torr = 58.19torr

Vapour pressure at 40°C(313 K) = 58.19 torr

Revisal Problems For JEE Advanced (Liquid Solutions) - Question 23

Passage l

W ater boils at 100°C at 1 atm pressure. Latent heat of vaporisation (ΔH ) = 10 kcal mol-1.

Q.

Lowering of vapour pressure of 0.2 molal urea solution at 40°C is

Detailed Solution for Revisal Problems For JEE Advanced (Liquid Solutions) - Question 23

Variation of vapour pressure with temperature is given Clapeyron-Clausius equation.

p1 = 0.07656 atm

p1 = 0.07656 x 760 torr = 58.19torr

Vapour pressure at 40°C(313 K) = 58.19 torr

Revisal Problems For JEE Advanced (Liquid Solutions) - Question 24

Vapour pressure diagram for real solutions of two liquids A and B that exhibit a negative deviation from Raoult’s law. The vapour pressure of both A and 8 are less than predicted by Raoult’s law. The dashed lines represent the plots for ideal solutions.

Q.

Solution containing components A and 8 shows this type of deviation from ideal behaviour when

Detailed Solution for Revisal Problems For JEE Advanced (Liquid Solutions) - Question 24

This is a case of negative deviation, thus force of attraction between (A ... B) is larger than average of (A .... A) and (B....B).

Thus, vapour pressure of mixture is decreased, boiling point is increased, hence

ΔH < 0, ΔV = negative

Diethylether and chloroform show negative deviation.

Revisal Problems For JEE Advanced (Liquid Solutions) - Question 25

Vapour pressure diagram for real solutions of two liquids A and B that exhibit a negative deviation from Raoult’s law. The vapour pressure of both A and 8 are less than predicted by Raoult’s law. The dashed lines represent the plots for ideal solutions.

Q.

This type of deviation is also expected in the following mixture.

Detailed Solution for Revisal Problems For JEE Advanced (Liquid Solutions) - Question 25

This is a case of negative deviation, thus force of attraction between (A ... B) is larger than average of (A ... A) and (B ... B). Thus, vapour pressure of mixture is decreased, boiling point is increased, hence

ΔH < 0, ΔV = negative

Diethylether and chloroform show negative deviation.

Revisal Problems For JEE Advanced (Liquid Solutions) - Question 26

Passage III

Vapour pressure of binary solution containing 6 g of non-volatile solute in 180 g water is 20.0 torr at 298 K. If one mole of water is further added, vapour pressure increases by 0.02 torr.

Q.

Molar mass of the solute is

Detailed Solution for Revisal Problems For JEE Advanced (Liquid Solutions) - Question 26

Revisal Problems For JEE Advanced (Liquid Solutions) - Question 27

Passage III

Vapour pressure of binary solution containing 6 g of non-volatile solute in 180 g water is 20.0 torr at 298 K. If one mole of water is further added, vapour pressure increases by 0.02 torr

Detailed Solution for Revisal Problems For JEE Advanced (Liquid Solutions) - Question 27

*Answer can only contain numeric values
Revisal Problems For JEE Advanced (Liquid Solutions) - Question 28

The so-called ozone layer in the earth's stratosphere has an average total pressure of 10 mm with partial pressure of ozone in the layer 1.2 x 10-5 mm. What is the concentration of ozone (in ppm) assuming that air is a mixture of N2 and 02 is 4 : 1 molar ratio.


Detailed Solution for Revisal Problems For JEE Advanced (Liquid Solutions) - Question 28

*Answer can only contain numeric values
Revisal Problems For JEE Advanced (Liquid Solutions) - Question 29

The vapour pressure of a solvent is decreased by 10 mm of Hg when a non-volatile solute was added to the solvent. The mole fraction of the solute in the solution is 0.2. How many times the vapour pressure is decreased when mole fraction of solvent is decreased by 25%?


Detailed Solution for Revisal Problems For JEE Advanced (Liquid Solutions) - Question 29

*Answer can only contain numeric values
Revisal Problems For JEE Advanced (Liquid Solutions) - Question 30

The freezing point depression of a 1.00x 10-3 molal aqueous solution of Kx[Fe(CN)6] is 7.1O x 10-3 K. If Kf (H2O)= 1.86 K kg mol-1, what is the value of x?


Detailed Solution for Revisal Problems For JEE Advanced (Liquid Solutions) - Question 30

∴    i = van't Hoff factor = 1 + ( y - 1 )α

where, α = degree of ionisation = 1 = 1 + (x + 1 — 1) 1 = (x + 1)

ΔTf = molality x Kf x i

7 x 10-3 = 1 x 10-3 x 1.86 x (x + 1)

x + 1 = 3.76 = 4

∴   x = 3

View more questions
Information about Revisal Problems For JEE Advanced (Liquid Solutions) Page
In this test you can find the Exam questions for Revisal Problems For JEE Advanced (Liquid Solutions) solved & explained in the simplest way possible. Besides giving Questions and answers for Revisal Problems For JEE Advanced (Liquid Solutions), EduRev gives you an ample number of Online tests for practice

Top Courses for JEE

Download as PDF

Top Courses for JEE