Q2:The plot of osmotic pressure (П) vs concentration (molL^{−1}) for a solution gives a straight line with slope 25.73L bar mol^{−1}. The temperature at which the osmotic pressure measurement is done is
(Use R = 0.083 L bar mol^{1} K^{1})
(a) 37^{∘}C
(b) 310^{∘}C
(c) 25.73^{∘}C
(d) 12.05^{∘}C (NEET 2024)
Ans: (a)
The relationship between the osmotic pressure (Π) of a solution and its concentration (c) can be derived from the van't Hoff equation for dilute solutions, which is given by:
Π = cRT
where:
Π is the osmotic pressure,
c is the concentration of the solution in moles per liter,
R is the ideal gas constant in appropriate units, and
T is the temperature in Kelvin.
According to the problem, the slope of the Π vs c plot is given as 25.73L bar mol^{−1} which corresponds to the product RT from the van't Hoff equation. We are provided with the value of the gas constant
R = 0.083 L bar mol^{−1} K^{−1}.
To find the temperature
T, we use the following equation derived from the slope of the line:
RT=25.73 L bar mol^{−}^{1}
To isolate T, we rearrange the equation:
T=25.73 L bar mol^{−1} / 0.083 L bar mol^{−1} K^{−1} ≈ 310 K
To convert this temperature from Kelvin to Celsius, we use the conversion formula:
This value is closest to 37^{∘}C, which corresponds to Option A. Therefore, the temperature at which the osmotic pressure measurement is done is approximately 37^{∘}C.
Q1: K_{H} value for some gases at the same temperature 'T' are given : (NEET 2022)
where K_{H} is Henry's Law constant in water. The order of their solubility in water is :
(a) HCHO < CH_{4} < CO_{2} < Ar
(b) Ar < CO_{2} < CH_{4} < HCHO
(c) Ar < CH_{4} < CO_{2} < HCHO
(d) HCHO < CO_{2} < CH_{4} < Ar
Ans: (b)
According to Henry's Law,
p = K_{H}x
Where 'p' is partial pressure of gas in vapour phase.
K_{H} is Henry's Law constant.
'x' is mole fraction of gas in liquid.
Higher the value of K_{H} at a given pressure, lower is the solubility of the gas in the liquid.
Q1: The following solutions were prepared by dissolving 10g of glucose (C_{6}H_{12}O_{6}) in 250 ml of water (P1). 10g of urea (CH_{4}N_{2}O) in 250 ml of water (P_{2}) and 10 g of sucrose (C_{12}H_{22}O_{11}) in 250 ml of water (P_{3}). The decreasing order of osmotic pressure of these solutions is: (NEET 2021)
(a) P_{2} > P_{3} > P_{1}
(b) P_{3} > P_{1} > P_{2}
(c) P_{2} > P_{1} > P_{3}
(d) P_{1} > P_{2 }> P_{3}
Ans: (c)
Q2: The correct option for the value of vapour pressure of a solution at 45°C with benzene to octane in a molar ratio 3 : 2 is: [At 45°C vapour pressure of benzene is 280 mm Hg and that of octane is 420 mm Hg. Assume Ideal gas] (NEET 2021)
(a) 336 mm of Hg
(b) 350 mm of Hg
(c) 160 mm of Hg
(d) 168 mm of Hg
Ans: (a)
Given, So,
Total vapour pressure of solution,
Q1: The mixture which shows positive deviation from Raoult’s law is : (NEET 2020)
(a) Acetone + Chloroform
(b) Chloroethane + Bromoethane
(c) Ethanol + Acetone
(d) Benzene + Toluene
Ans: (c)
Pure ethanol molecules are hydrogen bonded. On adding acetone, its molecules get in between the ethanol molecules and break some of the hydrogen bonds between them. This weakness the intermolecular attractive interactions and the solution shows positive deviation from Raoult's law.
Q2: The freezing point depression constant (K_{f}) of benzene is 5.12 K kg mol^{1}. The freezing point depression for the solution of molality 0.078 m containing a nonelectrolyte solute in benzene is (rounded off upto two decimal places) : (NEET 2020)
(a) 0.40 K
(b) 0.60 K
(c) 0.20 K
(d) 0.80 K
Ans: A
ΔT_{f} = ik_{f}m
⇒ ΔTf = 1 × 5.12 × 0.078
ΔT_{f} = 0.3993
ΔT_{f} = 0.40 K
Q1: For an ideal solution, the correct option is : (NEET 2019)
(a) Δ_{mix} S = 0 at constant T and P
(b) Δ_{mix} V ≠ 0 at constant T and P
(c) Δ_{mix} H = 0 at constant T and P
(d) Δ_{mix} G = 0 at constant T and P
Ans: (c)
For ideal solution,
Δ_{mix} H = 0
Δ_{mix} S > 0
Δ_{mix} G < 0
Δ_{mix} V = 0
Q2: The mixture that forms maximum boiling azeotrope is: (NEET 2019)
(a) Water + Nitric acid
(b) Ethanol + Water
(c) Acetone + Carbon disulphide
(d) Heptane + Octane
Ans: (a)
Maximum boiling azeotrope is shown by solution which shows negative deviation from Raoult's law. Except water + Nitric acid, all other mixtures show negative deviation.
Q1: If molality of the dilute solutions is doubled, the value of molal depression constant (K_{f}) will be : (NEET 2017)
(a) halved
(b) tripled
(c) unchanged
(d) doubled
Ans: (c)
The value of molal depression constant, K_{f }is constant for a particular solvent, thus, it will be unchanged when molality of the dilute solution is doubled.
Q1: Which one of the following is incorrect for ideal solution? (NEET 2016 Phase 2)
(a) Δ H_{mix} = 0
(b) Δ U_{mix} = 0
(c) ΔP = P_{obs}  P_{Calculated by }_{Raoult' Law}
_{(d) Δ Gmix = 0}
_{Ans: }_{(d)}
_{For ideal solution, we have}
_{Δ}_{Hmix = 0, ΔVmix = 0}
_{Now Umix = ΔHmix – PΔVmix}
_{$\therefore $ ΔUmix = 0}
_{Also, for an ideal solution,}
_{pA = xApAo, pB = xBpBo}
_{$\therefore $ Δp = pobserved – pcalculated = 0}
_{Δ}_{Gmix = ΔHmix – TΔSmix}
_{For an ideal solution, ΔSmix $\ne $ 0}
_{$\therefore $ ΔGmix $\ne $ 0 }
_{Q2: The van't Hoff factor (i) for a dilute aqueous solution of the strong electrolyte barium hydroxide is (NEET 2016 Phase 2)}
_{(a) 0}
_{(b) 1}
_{(c) 2}
_{(d) 3}
_{Ans: (d)}
_{Ba(OH)2 is a strong electrolyte and undergoes cent percent dissociation in a dilute aqueous solution.}
_{Ba(OH)2(aq) $\to $ Ba2+(aq) + 2OH–(aq)}
_{Thus, van’t Hoff factor i = 3. }
Q3: At 100°C the vapour pressure of a solution of 6.5 g of a solute in 100 g water is 732 mm. If K_{b} = 0.52, the boiling point of this solution will be : (NEET 2016 Phase 1)
(a) 103°C
(b) 101°C
(c) 100°C
(d) 102°C
Ans: (b)
Given that
w_{s} = 6.5 g, w_{A} = 100 g
p_{s} = 732 mm of Hg
k_{b} = 0.52, T^{o}_{b} = 100^{o}C
p^{o} = 760 mm of Hg
ΔT_{b} = K_{b} × m
Q4: Which of the following statements about the composition of the vapour over an ideal 1 : 1 molar mixture of benzene and toluene is correct?
Assume that the temperature is constant at 25°C. (Given Vapour Pressure Data at 25ºC, Benzene = 12.8kPa, toluene = 3.85kPa)
(a) Not enough information is given to make a prediction.
(b) The vapour will contain a higher percentage of benzene.
(c) The vapour will contain a higher percentage of toluene.The vapour will contain a higher percentage of toluene.
(d) The vapour will contain equal amounts of benzene and toluene. (NEET 2016 Phase 1)
Ans: (b)
p_{Benzene} = x_{Benzene}. p^{o}_{Benzene}
p_{Toluene} = x_{Toluene}. p^{o}_{Toluene}
For an ideal 1 : 1 molar mixture of benzene and toluene
x_{Benzene} =
p_{Benzene} =
p_{Toluene} =
Thus, the vapour will contain a high percentage of benzene as the partial vapour pressure of benzene is higher as compared to that of toluene.
Q1: Which one is not equal to zero for an ideal solution ? (NEET / AIPMT 2015 Cancelled Paper)
(a) ΔP = P_{observed}  P_{Raoult}
(b) Δ H_{mix}
(c) ΔS_{mix}
(d) Δ V_{mix}
Ans: (c)
For an ideal solution, ΔS_{mix} > 0 while ΔH_{mix}, ΔV_{mix} and ΔP all are 0.
Q2: The boiling point of 0.2 mol kg^{1} solution of X in water is greater than equimolal solution of Y in water. Which one of the following statements is true in this case ? (NEET / AIPMT 2015 Cancelled Paper)
(a) Y is undergoing dissociation in water while X under goes no change
(b) X is undergoing dissociation in water
(c) Molecular mass of X is greater than the molecular mass of Y.
(d) Molecular mass of X is less than the molecular mass of Y.
Ans: (b)
ΔT_{b} = iK_{b}m
Given, (ΔT_{b})_{x} > (ΔT_{b})_{y}
(K_{b} is same for same solvent)
So, x is undergoing dissociation in water.
Q3: Which one of the following electrolytes has the same value of van‘t Hoff‘s factor (i) as that of Al_{2}(SO_{4})_{3} (if all are 100% ionised) (NEET / AIPMT 2015 Cancelled Paper)
(a) K_{4}[Fe(CN)_{6}]
(b) K_{2}SO_{4}
(c) K_{3}[Fe(CN)_{6}]
(d) Al(NO_{3})_{3}
Ans: (a)
Ai_{2}(SO_{4})_{3} ⇌ 2Al^{+3} + 3SO_{4}^{2–}
van't Hoff factor, i = 5
K_{2}SO_{4} ⇌ 2K^{+} + SO_{4}^{2–}
van't Hoff factor, i = 3
K_{3}[Fe(CN)_{6}] ⇌ 3K^{+} + [Fe(CN)_{6}]^{2}
van't Hoff factor, i = 4
Al(NO_{3})_{3} ⇌ Al^{3+} + 3NO_{3}^{–}
van't Hoff factor, i = 4
K_{4}[Fe(CN)_{6}] ⇌ 4K^{+} + [Fe(CN)_{6}]^{4–}
van't Hoff factor, i = 5
Q1: Of the following 0.10 m aqueous solutions, which one will exhibit the largest freezing point depression? (NEET / AIPMT 2014)
(a) Al_{2} (SO_{4} )_{3}
(b) K_{2}SO_{4}
(c) KCl
(d) C_{6}H_{12}O_{6}
Ans: (a)
We know that depression in freezing point (ΔT_{f} ) is given as
ΔT_{f} = iK_{f}m
So, ΔT_{f}
Thus, more value of i (Van’t Hoff factor), more will be depression in freezing point.
Al_{2}(SO_{4})_{3} ⇌ 2Al^{+3} + 3SO_{4}^{2–}
i is maximum i.e., 5 for Al_{2}(SO_{4})_{3}
135 videos348 docs182 tests

1. What are the important topics covered in NEET exam from 2014 to 2024? 
2. How can I prepare effectively for NEET based on the previous year questions from 2014 to 2024? 
3. Are there any changes in the exam pattern of NEET from 2014 to 2024? 
4. How can solving previous year NEET questions help in boosting my preparation for the exam? 
5. Is it advisable to rely solely on previous year NEET questions for exam preparation? 

Explore Courses for NEET exam
