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JEE • CHEMISTRY • SOLUTIONS

Solutions practice · Mixed difficulty

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61 Hard
61 of 460
Application +4 | -1
Vapour pressure of chloroform (CHCl3)\left( {\mathrm{CHC}{\mathrm{l}_3}} \right) and dichloromethane (CH2Cl2)    at  25∘C\left( {\mathrm{C}{\mathrm{H}_2}\mathrm{C}{\mathrm{l}_2}} \right)\;\;at\;25^\circ C are 200 mm Hg and 41.5 mm Hg respectively. Vapour pressure of the solution obtained by mixing 25.5 g of (CHCl3)\left( {\mathrm{CHC}{\mathrm{l}_3}} \right) and 40 g of CH2Cl2{\mathrm{C}{\mathrm{H}_2}\mathrm{C}{\mathrm{l}_2}} at the same temperature will be :
(Molecular mass of CHCl3=119.5    u\mathrm{CHC}{\mathrm{l}_3} = 119.5\;\;u and molecular mass of CH2Cl2=85  u)\mathrm{C}{\mathrm{H}_2}\mathrm{C}{\mathrm{l}_2} = 85\;u)
View answer and explanation
Correct option: D
90.952 mm Hg
Option D (90.952 mm Hg) is correct because it's calculated using Raoult's Law, considering the mole fractions and vapor pressures of both components.
62 Easy
62 of 460
Understanding +4 | -1
A semipermeable membrane used in the measurement of osmotic pressure of a solution allows the passage of:
View answer and explanation
Correct option: A
solute molecules through it
Option B (solvent molecules through it) is correct because a semipermeable membrane allows only solvent molecules to pass through, preventing the passage of solute molecules.
63 Hard
63 of 460
Application +4 | -1
Mole fraction of solute in benzene is 0.2 then find molality of solute
View answer and explanation
Correct option: A
3.2
Option A (3.2) is correct because the molality is calculated using the mole fraction of the solute and the molar mass of benzene.
64 Medium
64 of 460
Application +4 | -1
A solution of urea boils at 100.180C{100.18^0}C at the atmospheric pressure. If Kf{K_f} and Kb{K_b} for water are 1.86 and 0.512 K.Kg.Mol−1K.Kg.Mo{l^{ - 1}} respectively, the above solution will freeze at
View answer and explanation
Correct option: B
−0.654 0C- 0.654{\,^0}C
Option B (-0.654 °C) is correct because the freezing point depression can be calculated using the boiling point elevation and the cryoscopic and ebullioscopic constants.
65 Hard
65 of 460
Application +4 | -1
In a Victor Meyer determination of the relative molecular mass of benzene, the heating vessel was maintained at 120∘C.120^{\circ} \mathrm{C}. A mass of 0.1528 g0.1528 \mathrm{~g} of benzene was used and the volume of displaced air collected over water at 15∘C,15^{\circ} \mathrm{C}, was 48 cm3.48 \mathrm{~cm}^3. The barometric pressure was 743 mm Hg.743 \mathrm{~mm} \mathrm{~Hg}. Calculate the relative molecular mass of benzene. The vapor pressure of water at 15∘C15^{\circ} \mathrm{C} is 13 mm Hg.13 \mathrm{~mm} \mathrm{~Hg}.
View answer and explanation
Correct option: B
78.26
Option B (78.26) is correct because it is calculated using the ideal gas law and accounting for the vapor pressure of water.
66 Easy
66 of 460
Understanding +4 | -1
Which of the following is not an example of a solution?
View answer and explanation
Correct option: D
Benzene in water
Option D (Benzene in water) is correct because benzene and water are immiscible, forming a heterogeneous mixture, not a homogeneous solution.
67 Medium
67 of 460
Application +4 | -1
An aqueous solution of a weak monobasic acid containing 0.1g0.1 g in 21.7g21.7 g of water freezes at 272.813 K272.813\, K. If the value of KfK_f for water is 1.86 K/m1.86 {~K} / {m}, what is the molecular mass of the monobic acid
View answer and explanation
Correct option: D
60 g/mole60 {~g} / {mole}
The correct answer is D (60 g/mole) because the molecular mass is calculated using the freezing point depression formula.
68 Medium
68 of 460
Application +4 | -1
When 45 gm solute is dissolved in 600 gm water freezing point lower by 2.2 K, calculate molar mass of solute (Kf=1.86 K kg)\left( {{K_f} = 1.86\,K\,kg} \right)
View answer and explanation
Correct option: A
63.4
Option A (63.4) is correct because the molar mass of the solute is calculated using the freezing point depression formula.
69 Hard
69 of 460
Analyzing +4 | -1
A non-volatile solute 'X\mathrm{X}' completely dimerizes in water, if the temperature is below −3.72∘C-3.72^{\circ} \mathrm{C} and the solute completely dissociates as X→Y+Z,\mathrm{X} \rightarrow \mathrm{Y}+\mathrm{Z}, if the temperature is above 100.26∘C.100.26^{\circ} \mathrm{C}. In between these two temperatures (including both temperatures), the solute is neither dissociated nor associated. One mole of 'X\mathrm{X}' is dissolved in 1.0 kg1.0 \mathrm{~kg} water (Kb=0.52 K−kg/mol,Kf=1.86 K−kg/mol)\left(K_{\mathrm{b}}=0.52 \mathrm{~K}-\mathrm{kg} / \mathrm{mol}, K_{\mathrm{f}}=1.86 \mathrm{~K}-\mathrm{kg} / \mathrm{mol}\right). Identify the incorrect information related with the solution.
View answer and explanation
Correct option: C
When the solution is cooled to −7.44∘C,75%-7.44^{\circ} \mathrm{C}, 75 \% of water present initially will separate as ice
Option D is incorrect because the percentage of water separated as ice when cooled to −7.44∘C-7.44^{\circ} \mathrm{C} is calculated to be 87.5%87.5\%, not 50%50\%.
70 Medium
70 of 460
Understanding +4 | -1
Assertion (A): People taking a lot of salty food experience the puffiness or swelling, called edema.

Reason (R): There is water retention in tissue cells and intercellular spaces hecause of osmosis.
View answer and explanation
Correct option: A
Both A\mathrm{A} and R\mathrm{R} are correct, R\mathrm{R} is the correct explanation of A\mathrm{A}.
Option A (Both A and R are correct, R is the correct explanation of A) is correct because salty food increases the solute concentration in the body, causing water retention in tissues due to osmosis, leading to edema.
71 Easy
71 of 460
Application +4 | -1
How many gram of glucose must be present in 0.5 L0.5 \mathrm{~L} of solution for its osmotic pressure to be the same as that of 9.2 g9.2 \mathrm{~g} of glucose per litre?
View answer and explanation
Correct option: A
4.6 g4.6 \mathrm{~g}
Option A (4.6 g4.6 \mathrm{~g}) is correct because for isotonic solutions, the ratio of mass to molar mass and volume must be equal.
72 Medium
72 of 460
Application +4 | -1
Consider the following arrangement, in which a solution containing 20 g20 \mathrm{~g} of haemoglobin in 1 dm31 \mathrm{~dm}^3 of the solution is placed in right compartment and pure water is placed in left compartment, separated by SPM. At equilibrium, the height of liquid in the right compartment is 74.5 mm74.5 \mathrm{~mm} in excess of that in the left compartment. The temperature of the system is maintained at 298 K.298 \mathrm{~K}. The number of millimoles in 320 g320 \mathrm{~g} of haemoglobin is (Given: The density of final solution is 1.013 g/ml,g=10 m/s21.013 \mathrm{~g} / \mathrm{ml}, \mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2, R=0.08 L−atm/K−molR=0.08 \mathrm{~L}-\mathrm{atm} / \mathrm{K}-\mathrm{mol})
View answer and explanation
Correct option: A
5
Option A (5) is correct because the number of millimoles is calculated using the osmotic pressure equation and the given parameters.
73 Hard
73 of 460
Remembering +4 | -1
Scuba divers may experience a condition called To avoids this, the tanks used by scuba divers are filled with air diluted with
View answer and explanation
Correct option: D
Bends, Helium
Option D (Bends, Helium) is correct because scuba divers use helium-diluted air to avoid bends caused by nitrogen bubbles in the blood.
74 Medium
74 of 460
Application +4 | -1
If the elevation in boiling point of a solution of 10 g of solute (mol. wt. = 100 g of water is ΔTb\Delta {T_b}, the ebullioscopic constant of water is
View answer and explanation
Correct option: C
ΔTb\Delta {T_b}
Option C (ΔTb\Delta {T_b}) is correct because the ebullioscopic constant is equal to the elevation in boiling point under the given conditions.
75 Medium
75 of 460
Application +4 | -1
2.5 g2.5 {~g} of a substance is present in 200 mL200 {~mL} of solution showing the osmotic pressure of 60 cmHg60 {~cm} {Hg} at 15∘C15^{\circ} {C}. Calculate the molar mass of substance. What will be the osmotic pressure if temperature is raised to 25∘C25^{\circ} {C} ?
View answer and explanation
Correct option: A
374.38
Option A (374.38) is correct because it is calculated using the osmotic pressure formula and the ideal gas constant.
76 Medium
76 of 460
Analyzing +4 | -1
Each of the following pair shows a positive deviation from the Raoult's law except
View answer and explanation
Correct option: A
(C2H5)O+HCl\left({C}_2 {H}_5\right) {O}+{HCl}
Option A ((C2H5)O+HCl\left({C}_2 {H}_5\right) {O}+{HCl}) is correct because ether and HCl form hydrogen bonds, leading to stronger interactions and a negative deviation from Raoult's law.
77 Medium
77 of 460
Understanding +4 | -1
Consider the solution of ethanol. In it
View answer and explanation
Correct option: D
Both (1) and (2)
Option D (Both (1) and (2)) is correct because in a solution of ethanol, ethanol is dissolved in water and both components are in the same phase (liquid).
78 Medium
78 of 460
Analyzing +4 | -1
Select correct statements :
View answer and explanation
Correct option: D
All are correct statements
The correct answer is D (All are correct statements) because all the statements regarding colligative properties, freezing point depression, and isotonic solutions are accurate.
79 Medium
79 of 460
Analyzing +4 | -1
Statement I: Mass density of a mixture of 20 g20 \mathrm{~g} water and 100 g100 \mathrm{~g} of ethanol is 0.86 g cm−30.86 \mathrm{~g} \mathrm{~cm}^{-3} at 298 K298 \mathrm{~K}.

Statement II: At 298 K298 \mathrm{~K}, molar volume of water and ethanol are respectively 18 cm318 \mathrm{~cm}^3 and 55 cm355 \mathrm{~cm}^3 based on density. (d=1.00 g cm−3\left(d=1.00 \mathrm{~g} \mathrm{~cm}^{-3}\right. for water and d=0.8364 g cm−3d=0.8364 \mathrm{~g} \mathrm{~cm}^{-3} for ethanol).
View answer and explanation
Correct option: B
Both Statement I and Statement II are correct but Statement II is not the correct explanation of Statement I.
Option B (Both Statement I and Statement II are correct but Statement II is not the correct explanation of Statement I) is correct because both statements are factually correct, but the molar volumes do not directly explain the mixture's density.
80 Medium
80 of 460
Understanding +4 | -1
Statement I: Normality and molarity can be calculated from each other.

Statement II: Normality is equal to the product of molarity and n\mathrm{n}.
View answer and explanation
Correct option: A
If both Statement I and Statement II are correct and Statement II is the correct explanation of Statement I
Option A is correct because Normality is defined as the product of Molarity and the n-factor (number of equivalents), making the statements correct and related.
81 Medium
81 of 460
Application +4 | -1
An aqueous solution of a non-volatile and non-electrolyte solute (molecular mass = 150) boils at 373.26 K.373.26 \mathrm{~K}. The composition of solution, in terms of mass percent of the solute, is (Kb\left(K_{\mathrm{b}}\right. of water == 0.52)0.52)
View answer and explanation
Correct option: C
6.98%6.98 \%
Option C (6.98 %) is correct because it uses the elevation in boiling point formula to find the molality and then calculates the mass percent of the solute.
82 Medium
82 of 460
Application +4 | -1
Which of the following has lowest freezing point
View answer and explanation
Correct option: B
0.1 M0.1 \,{M} aqueous solution of NaCl{NaCl}
Option B (5.6 V) is correct because the volume strength of 1N H2O2 is 5.6 V, which is a direct application of the formula.
83 Medium
83 of 460
Understanding +4 | -1
A perfectly semipermeable membrane when used to separate a solution from its solvent permits through it the passage of:
View answer and explanation
Correct option: B
Solvent only
Option B (Solvent only) is correct because a perfectly semipermeable membrane allows only solvent molecules to pass through, blocking solute molecules.
84 Medium
84 of 460
Understanding +4 | -1
At higher altitudes the boiling point of water lowers because
View answer and explanation
Correct option: A
Atmospheric pressure is low
Option A is correct because at higher altitudes, the atmospheric pressure is lower, which causes water to boil at a lower temperature.
85 Easy
85 of 460
Application +4 | -1
Which of the following shows maximum depression in freezing point?
View answer and explanation
Correct option: A
K2SO4{K_2}S{O_4}
Option A (K2SO4) is correct because it dissociates into the highest number of ions (3), leading to the maximum depression in freezing point, a colligative property.
86 Hard
86 of 460
Application +4 | -1
Assuming ideal behaviour, what mass of a non-volatile solid of molar mass 128 g mole−1128 \mathrm{~g} \mathrm{~mole}^{-1} would have to be dissolved in 228 g228 \mathrm{~g} octane (CgH18)\left(\mathrm{C}_{\mathrm{g}} \mathrm{H}_{18}\right) to lower the vapour pressure of octane by 20%20 \%?
View answer and explanation
Correct option: A
64 g64 \mathrm{~g}
Option A (64 g64 \mathrm{~g}) is correct because it is calculated using Raoult's law, considering the molar masses and the given vapor pressure reduction.
87 Hard
87 of 460
Application +4 | -1
An ideal solution made by mixing two volatile components A and B. If the vapour pressure of solution is 170 mm Hg, then mole fraction of A in vapour phase is (PA0=200 mm Hg, PB0=100 mm Hg )\left( {P_A^0 = 200\,mm\,Hg,\,P_B^0 = 100\,mm\,Hg\,} \right)
View answer and explanation
Correct option: B
0.82
Option B (0.82) is correct because it is calculated using Raoult's law and Dalton's law to determine the mole fraction of A in the vapor phase.
88 Medium
88 of 460
Understanding +4 | -1
Which one of the following gases has the lowest value of Henry's law constant?
View answer and explanation
Correct option: D
CO2\mathrm{CO}_2
Option D (CO2\mathrm{CO}_2) is correct because carbon dioxide has the highest solubility in water among the given gases, leading to the lowest Henry's law constant.
89 Medium
89 of 460
Analyzing +4 | -1
An open beaker of water in equilibrium with water vapour is in a sealed container. When few grams of glucose are added to the beaker of water, the rate at which water molecules
View answer and explanation
Correct option: C
leaves the vapour increases
Option C is correct because adding glucose to water lowers the water's vapor pressure, causing more water to evaporate to re-establish equilibrium.
90 Medium
90 of 460
Application +4 | -1
One mole of a solute A\mathrm{A} is dissolved in a given volume of a solvent. The association of the solute take place as follows: nA⇌An\mathrm{nA} \rightleftharpoons \mathrm{A}_{\mathrm{n}}. If α\alpha is the degree of association of A, the van't Hoff factor ' i\mathrm{i} ' is expressed as:
View answer and explanation
Correct option: C
i=1−α+αn1\mathrm{i}=\frac{1-\alpha+\frac{\alpha}{\mathrm{n}}}{1}
Option C (6 min) is correct because it is the half-life calculated from the rate constants at two different times for a first-order reaction.
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