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JEE • CHEMISTRY • SOLUTIONS

Solutions practice · Medium

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These sets use single-correct-answer MCQs. They do not cover every JEE Main or JEE Advanced question format and are not full-length mock papers.

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Showing 211-240 of 265 questions
211 Medium
211 of 265
Application +4 | -1
The ratio of elevation in boiling point for Fe3[Fe(CN)6]2{Fe}_3\left[{Fe}({CN})_6\right]_2 (assume complete ionization) to elevation in boiling point of sugar solution of equal molal concentration is
View answer and explanation
Correct option: A
5:15: 1
The correct answer is A (5:15: 1) because the ratio of elevation in boiling point is directly proportional to the van't Hoff factor, which is 5 for the complex and 1 for sugar.
212 Medium
212 of 265
Understanding +4 | -1
Statement I: Vapour pressure of water is less than 1.013 bar at 373 K373 \mathrm{~K}.

Statement II: Water boils at 373 K373 \mathrm{~K} as the vapour pressure of water at this temperature becomes equal to atmospheric pressure
View answer and explanation
Correct option: D
If Statement I is false but Statement II is correct
Option D (If Statement I is false but Statement II is correct) is correct because the vapor pressure of water at 373 K is equal to 1.013 bar, and water boils when its vapor pressure equals atmospheric pressure.
213 Medium
213 of 265
Analyzing +4 | -1
Assertion (A): 0.1 M solution of glucose has higher increment in the freezing point than 0.1 M solution of urea.

Reason (R): Kf\mathrm{K}_{\mathrm{f}} for both has different values.
View answer and explanation
Correct option: D
Both (A) and (R) are false
Option D (Both (A) and (R) are false) is correct because the freezing point depression is a colligative property and depends on the number of particles. Glucose and urea are both non-electrolytes, so for the same molarity, they will have the same freezing point depression. Also, Kf is a constant for a given solvent.
214 Medium
214 of 265
Application +4 | -1
Find out molality of 0.8 M H2SO40.8 \mathrm{~M} \mathrm{~H}_2 \mathrm{SO}_4 solution having density of solution equal to 1.02 gm/ml1.02 \mathrm{~gm} / \mathrm{ml} (Nearest integer)
View answer and explanation
Correct option: A
1
Option A (1) is correct because the molality is calculated using the given molarity and density, and the result is closest to 1.
215 Medium
215 of 265
Understanding +4 | -1
Assertion (A) : At low concentration, benzene and toluene forms ideal solution.

Reason (R) : Components with structural similarity forms ideal solution
View answer and explanation
Correct option: C
Both Assertion and Reason are correct and Reason is correct explantion of Assertion
Option C is correct because benzene and toluene are structurally similar and form ideal solutions at low concentrations, and structural similarity is the reason for ideal solution formation.
216 Medium
216 of 265
Application +4 | -1
Ice begins to separate at −0.744∘C-0.744^{\circ} \mathrm{C} from an aqueous solution containing 2.4 g2.4 \mathrm{~g} of a non-volatile, non-electrolyte in 100 g100 \mathrm{~g} water. The osmotic pressure of the solution at 27∘C27^{\circ} \mathrm{C} is (Kf\left(K_{\mathrm{f}}\right. for water =1.86,=1.86, assume molality == molarity)
View answer and explanation
Correct option: B
9.852 atm
Option B (9.852 atm) is correct because the osmotic pressure is calculated using the freezing point depression and the ideal gas law.
217 Medium
217 of 265
Analyzing +4 | -1
Statement I: The freezing point of water is depressed by the addition of glucose.

Statement II: Entropy of solution is less than entropy of pure solvent.
View answer and explanation
Correct option: C
If Statement I is correct but Statement II is false
Statement I is correct because the addition of a solute like glucose lowers the freezing point of water. Statement II is incorrect because the entropy of a solution is greater than that of a pure solvent. Therefore, the correct answer is C.
218 Medium
218 of 265
Analyzing +4 | -1
Consider following terms (m=(\mathrm{m}= molality )) :

I : mKb;{mK}_{{b}};\quad II: mKbi{mK}_{{b}} {i}\quad III : ΔTbi\frac{\Delta {T}_{{b}}}{{i}}\quad IV : KbK_b

Terms which can be expressed in degree (temperature) are
View answer and explanation
Correct option: C
I, II, IIII
Option D (I, III) is correct because mKb{mK}_{{b}} and ΔTbi\frac{\Delta {T}_{{b}}}{{i}} can be expressed in units of temperature.
219 Medium
219 of 265
Understanding +4 | -1
The degree of dissociation (∝\propto) of a weak electrolyte, AxBy{A_x}By is related to vant Hoff factor (i) by the expression
View answer and explanation
Correct option: C
∝=i−1x+y−1\propto = \frac{{i - 1}}{{x + y - 1}}
Option C (∝=i−1x+y−1\propto = \frac{{i - 1}}{{x + y - 1}}) is correct because it represents the correct relationship between the degree of dissociation (α\alpha) and the van't Hoff factor (i) for a weak electrolyte.
220 Medium
220 of 265
Application +4 | -1
50 g of an acid is dissolved in one litre aqueous solution. Distribution coefficient in favour of ether is 3. Acid left in aqueous layer when solution is shaken with one litre ether :
View answer and explanation
Correct option: B
12.5 g
Option B (12.5 g) is correct because it represents the amount of acid left in the aqueous layer after distribution, calculated using the distribution coefficient.
221 Medium
221 of 265
Analyzing +4 | -1
The solution from the following with highest depression in freezing point/lowest freezing point is
View answer and explanation
Correct option: B
180 g180 \mathrm{~g} of acetic acid dissolved in water
Option B (180 g180 \mathrm{~g} of acetic acid dissolved in water) is correct because acetic acid dissociates in water, leading to a higher number of solute particles compared to the other options, thus resulting in the highest depression in freezing point.
222 Medium
222 of 265
Understanding +4 | -1
Select correct statement(s) :
View answer and explanation
Correct option: C
Both (1) and (2)
Option C (Both (1) and (2)) is correct because adding a non-volatile solute elevates the boiling point and depresses the freezing point.
223 Medium
223 of 265
Application +4 | -1
0.1435 m0.1435 \mathrm{~m} solution of a non-volatile, non-electrolyte solute has the freezing point 0.73 degrees lower than that of benzene. What is the value of molal freezing point depression constant of benzene?
View answer and explanation
Correct option: A
5.087 Km−15.087 \mathrm{~Km}^{-1}
Option A (5.087 Km−15.087 \mathrm{~Km}^{-1}) is correct because the molal freezing point depression constant is calculated using the formula ΔTf=Kf⋅m\Delta T_f = K_f \cdot m.
224 Medium
224 of 265
Analyzing +4 | -1
The plots of 1XA  vs.  1YA\frac{1}{{{X_A}}}\;vs.\;\frac{1}{{{Y_A}}} (where XA  and  YA{X_A}\;\mathrm{and}\;{Y_A} are the mole fraction of liquid A in liquid and vapour phase respectively) is linear with slope and intercepts respectively:
View answer and explanation
Correct option: B
PA∘/PB∘  and  (PB∘−PA∘)PB∘P_A^ \circ /P_B^ \circ \;and\;\frac{{(P_B^ \circ - P_A^ \circ )}}{{P_B^ \circ }}
Option D (PB∘/PA∘  and  (PB∘−PA∘)PB∘P_B^ \circ /P_A^ \circ \;and\;\frac{{(P_B^ \circ - P_A^ \circ )}}{{P_B^ \circ }}) is correct because it matches the derived slope and intercept from Raoult's law and Dalton's law.
225 Medium
225 of 265
Application +4 | -1
One gram of silver gets distributed between 10 cm310{\kern 1pt} \,\mathrm{c}{\mathrm{m}^3} of molten zinc and 100 cm3100\,\mathrm{c}{\mathrm{m}^3} of molten lead at 8000∘C.8000^\circ C. The percentage of silver still left in the lead layer in approximately
View answer and explanation
Correct option: C
3
Option C (3) is correct because, using the distribution coefficient formula, the percentage of silver left in the lead layer is approximately 3%.
226 Medium
226 of 265
Application +4 | -1
18 g of Glucose is added to 178.2 g of water the vapour pressure of water for this aqeous solution at 100∘C100^\circ C is.
View answer and explanation
Correct option: A
752.4 torr
Option A (752.4 torr) is correct because it is calculated using Raoult's Law, considering the mole fraction of water in the solution.
227 Medium
227 of 265
Understanding +4 | -1
Assertion (A): Sum of mole fraction of all components in a mixture is 1.

Reason (R): Mole fraction is temperature dependent mode of concentration.
View answer and explanation
Correct option: C
A is true, R is false
Option C is correct because the sum of mole fractions of all components in a mixture is always 1, but mole fraction is independent of temperature.
228 Medium
228 of 265
Application +4 | -1
The osmatic pressure of a 5% (wt./vol) solution of cane sugar at 150∘C150^\circ C is
View answer and explanation
Correct option: C
5.078  atm5.078\;\mathrm{ }atm
Option C (5.078  atm5.078\;\mathrm{ }atm) is correct because it is calculated using the osmotic pressure formula π=CRT\pi = CRT.
229 Medium
229 of 265
Understanding +4 | -1
Assertion (A) : The pressure exerted by a liquid in equilibrium with its liquid at a given temperature is called its vapour pressure.

Reason (R) : If a non-volatile solute is added to a solvent to give a solution, the vapour pressure of the solution is found to be greater than the vapour pressure of the pure solvent.
View answer and explanation
Correct option: A
Assertion is correct but Reason is Wrong
Option A is correct because the assertion is correct (vapor pressure definition), but the reason is wrong (adding a non-volatile solute lowers the vapor pressure).
230 Medium
230 of 265
Understanding +4 | -1
Which of the following is incorrect regarding solution showing negative deviation from Raoult's law?
View answer and explanation
Correct option: D
Δmix S<0\Delta_{\text {mix }} S < 0
Option D (Δmix S<0\Delta_{\text {mix }} S < 0) is correct because for solutions, the entropy of mixing is always positive (Δmix S>0\Delta_{\text {mix }} S > 0). Negative deviation does not change this fundamental property.
231 Medium
231 of 265
Application +4 | -1
The increase in boiling point of a solution containing 0.6 g urea in 200 g water is 0.50∘C0.50^\circ \mathrm{C} .Find the molal elevation constant.
View answer and explanation
Correct option: A
10  K  kg  mol−110\;K\;kg\;mo{l^{ - 1}}
Option D (1.0Kkgmol−11.0 K kg mol^{-1}) is correct because it is calculated using the formula for elevation in boiling point, considering the mass of urea, mass of water, and the molar mass of urea.
232 Medium
232 of 265
Application +4 | -1
What is the molality of a solution which contains 18 g18 {~g} of glucose (C6H12O6)\left({C}_6 {H}_{12} {O}_6\right) in 250 g250 {~g} of water
View answer and explanation
Correct option: B
0.4 m0.4 {~m}
Option B (0.4 m0.4 {~m}) is correct because molality is calculated by dividing the moles of solute by the kilograms of solvent.
233 Medium
233 of 265
Application +4 | -1
Boiling point of chloroform was raised by 0.323 K0.323 \mathrm{~K}, when 0.5143 g0.5143 \mathrm{~g} of anthracence was dissolved in its 35 g35 \mathrm{~g}. Molecular mass of anthracence is (Kb\left({K}_{{b}}\right. for CHCl3=3.9 k−kg−mol−1)\left.{CHCl}_3=3.9 \, \mathrm{k}-\mathrm{kg}-\mathrm{mol}^{-1}\right)
View answer and explanation
Correct option: C
177.42 g/mol177.42 \mathrm{~g} / \mathrm{mol}
Option C (177.42 g/mol) is correct because it is the molecular mass of anthracene calculated using the elevation in boiling point formula.
234 Medium
234 of 265
Application +4 | -1
An ideal solution was found to have a vapour pressure of 80 torr when the mole fraction of a non-volatile solute was 0.2.0.2. What would be the vapour pressure of the pure solvent at the same temperature?
View answer and explanation
Correct option: A
100
Option A (100) is correct because it is calculated using Raoult's law for ideal solutions, relating vapor pressure lowering to the mole fraction of the non-volatile solute.
235 Medium
235 of 265
Application +4 | -1
The solution containing 4.0 g of PVC in 1 L of
dioxane was found to have osmotic pressure of
0.006 atm at 300 K. The molecular mass of the
polymer PVC is
View answer and explanation
Correct option: A
16420
Option A (16,420) is correct because it is calculated using the osmotic pressure formula, solving for the molecular mass of the polymer.
236 Medium
236 of 265
Analyzing +4 | -1
1.0 g1.0 \mathrm{~g} of a mono basic acid HA\mathrm{HA} in 100 gm100 \mathrm{~gm} water lowers the freezing point by 0.155 K0.155 \mathrm{~K}. If 0.75 gm of same acid requires 25ml25 \mathrm{ml} of N/5NaOH\mathrm{N} / 5 \mathrm{NaOH} solution for complete neutralization then % degree of ionization of acid is (Kf(H2O)=1.86 kg mol−1)\left(K_f\left(H_2 \mathrm{O}\right)=1.86 \mathrm{~kg} \mathrm{~mol}^{-1}\right)
View answer and explanation
Correct option: A
25
The correct answer is A (25) because the degree of ionization is calculated by comparing the observed and normal molar masses of the acid, using freezing point depression and neutralization data.
237 Medium
237 of 265
Analyzing +4 | -1
Which pair will not be isotonic
View answer and explanation
Correct option: D
2.5 M−KCl  2.5\,M - KCl\,\, and 1M−K3[Fe(CN)6]1M - {K_3}\left[ {Fe{{\left( {CN} \right)}_6}} \right]
Option D is correct because 2.5 M KCl yields 5.0 M ions, while 1 M K3[Fe(CN)6] yields 4.0 M ions, thus they are not isotonic.
238 Medium
238 of 265
Application +4 | -1
What will be the molarity of a 200 mL200 \mathrm{~mL} solution containing 1.2 g1.2 \mathrm{~g} of urea?
View answer and explanation
Correct option: C
0.1 M0.1 \mathrm{~M}
Option C (0.1 M) is correct because molarity is calculated by dividing moles of solute by volume of solution in liters.
239 Medium
239 of 265
Application +4 | -1
If 0.1 M0.1 \mathrm{~M} solution of NaCl{NaCl} has same osmotic pressure as 0.1 M0.1 \mathrm{~M} solution of CaCl2,{CaCl}_2, then the degree of dissociation of CaCl2{CaCl}_2 will be
View answer and explanation
Correct option: A
0.5
Option A (0.5) is correct because the degree of dissociation is calculated by equating the osmolarities of the two solutions.
240 Medium
240 of 265
Application +4 | -1
The amount of solute (molar mass 60 g mol−160 \mathrm{~g} \mathrm{~mol}^{-1}) that must be added to 180 g180 \mathrm{~g} of water so that the vapour pressure of water is lowered by 10%10 \% is
View answer and explanation
Correct option: A
60 g60 \mathrm{~g}
Option A (60 g) is correct because it calculates the mass of solute needed to lower the vapor pressure by 10% using Raoult's Law.
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