31 Years NEET Previous Year Questions: Solutions - 1 Free MCQ Practice

Boiling point elevation is a colligative property:

ΔT_b = i K_b m

For the same solvent (water), K_b is constant, and at these low concentrations we can compare i×molality (≈ molarity).

• Na₂SO₄ → dissociates ideally into 3 ions (2 Na⁺ + SO₄^2-), so i ≈ 3
  i × M = 3 × 0.01 = 0.03

• KNO₃ → 2 ions, i ≈ 2
  2 × 0.01 = 0.02

• Glucose (C₆H₁₂O₆) → non-electrolyte, i = 1
  1×0.015=0.015

• Urea → non-electrolyte, i=1i=1
  1×0.01=0.011×0.01=0.01

Largest iM → Na₂SO₄, so it gives the highest boiling point.

• A. Humidity: Humidity refers to water vapor (liquid) dispersed in air (gas). This is an example of a liquid in gas solution. (A-II)
• B. Alloys: Alloys are homogeneous mixtures of metals or a metal and a non-metal in solid form. This is an example of a solid in solid solution. (B-I)
• C. Amalgams: Amalgams are alloys that consist of a metal (solid) dissolved in mercury (liquid). This is an example of a liquid in solid solution. (C-IV)
• D. Smoke: Smoke consists of tiny solid particles dispersed in a gas. This is an example of a solid in gas solution. (D-III)

Therefore, the correct option is (d) A-II, B-I, C-IV, D-III. 

Given:

• n_X = 5
• n_Y = 10
• P_X⁰ = 63 torr
• P_Y⁰ = 78 torr
• Observed total pressure = 70 torr

1. Mole Fractions:

The total number of moles is:

n_total = n_X + n_Y = 5 + 10 = 15

The mole fraction of X and Y are calculated as follows:

χ_X = χ_X / n_total = 5 / 15 = 0.333

χ_Y = χ_Y / n_total = 10 / 15 = 0.667

2. Predicted Vapor Pressure (Raoult's Law, Ideal Solution):

Using Raoult’s Law:

P_ideal = χ_X P_X⁰ + χ_Y P_Y⁰

Substitute the values:

P_ideal = (0.333)(63) + (0.667)(78)

First term:

(0.333)(63) = 20.999 ≈ 21

Second term:

(0.667)(78) = 52.026 ≈ 52

Thus:

P_ideal = 21 + 52 = 73 torr

3. Comparison with Observed:

• Predicted (ideal) pressure = 73 torr
• Observed pressure = 70 torr

Since P_obs < P_ideal, the solution shows a negative deviation from Raoult’s law.

According to Henry's law, the solubility of a gas in a liquid under constant temperature is directly proportional to the partial pressure of the gas above the liquid but is inversely proportional to the Henry's law constant (K_H) for the gas. Henry's law can be expressed as:

where c is the concentration (or solubility) of the gas in the liquid,
P is the partial pressure of the gas, and K_H is Henry's law constant.
From the equation, it is clear that the solubility of the gas is inversely related to
K_H; if K_H increases, the solubility decreases, and vice versa.
In the given problem, you have the K_H values of the gases A, B, and C as follows:
A: 145 kbar
B: 2×10⁻⁵ kbar
C: 35 kbar
Comparing the K_H values:
Gas B has the lowest K_H and hence the highest solubility.
Gas A, with the highest K_H among the three, will have the lowest solubility.
Gas C has a K_H value less than A but greater than B, so its solubility will be lower than B but higher than A.
Therefore, the order of solubility of the gases in water from the highest to the lowest is:
B > C > A\
Thus, the correct option is:
Option B: B > C > 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.

Osmotic pressure depends on the van't Hoff factor (i), which represents the number of particles a solute dissociates into, and the molarity (M) of the solution.
For two solutions to have the same osmotic pressure, the product of i and M should be the same.
Let's go through the options:

(a) 0.1 M BaCl₂ and 0.2 M K₂SO₄:

• BaCl₂ dissociates into 3 ions, so i = 3.
• K₂SO₄ dissociates into 3 ions, so i = 3.
• The product of i and M for BaCl₂ = 0.1 × 3 = 0.3.
• The product of i and M for K₂SO₄ = 0.2 × 3 = 0.6.
• These values are not equal, so they do not have the same osmotic pressure.

(b) 0.1 M Na₃PO₄ and 0.1 M K₂SO₄:

• Na₃PO₄ dissociates into 4 ions, so i = 4.
• K₂SO₄ dissociates into 3 ions, so i = 3.
• The product of i and M for Na₃PO₄ = 0.1 × 4 = 0.4.
• The product of i and M for K₂SO₄ = 0.1 × 3 = 0.3.
• These values are not equal, so they do not have the same osmotic pressure.

(c) 0.2 M NaCl and 0.1 M K₂SO₄:

• NaCl dissociates into 2 ions, so i = 2.
• K₂SO₄ dissociates into 3 ions, so i = 3.
• The product of i and M for NaCl = 0.2 × 2 = 0.4.
• The product of i and M for K₂SO₄ = 0.1 × 3 = 0.3.
• These values are not equal, so they do not have the same osmotic pressure.

(d) 0.2 M NaCl and 0.1 M K₃[Fe(CN)₆]:

• NaCl dissociates into 2 ions, so i = 2.
• K₃[Fe(CN)₆] dissociates into 4 ions, so i = 4.
• The product of i and M for NaCl = 0.2 × 2 = 0.4.
• The product of i and M for K₃[Fe(CN)₆] = 0.1 × 4 = 0.4.
• These values are equal, so they have the same osmotic pressure.

Thus, the correct answer is (d) 0.2 M NaCl and 0.1 M K₃[Fe(CN)₆].

• A. Liquid ⇌ Vapour: This process refers to boiling point, which is the temperature at which the liquid changes to vapor. So, A → II (Boiling point).
• B. Solid ⇌ Liquid: This process is the melting point, the temperature at which a solid changes to a liquid. So, B → I (Melting point).
• C. Solid ⇌ Vapour: This process is known as sublimation, where a solid directly changes into a vapor without passing through the liquid phase. So, C → III (Sublimation point).
• D. Solute(solid) ⇌ Solute(solution): This refers to the formation of a saturated solution, where the maximum amount of solute is dissolved in the solvent at a particular temperature. So, D → IV (Saturated solution).

Thus, the correct answer is A-II, B-I, C-III, D-IV.

To find the molar mass of the solute, we will use the formula for the depression in freezing point:
ΔT_f = K_f × m
Where:

• ΔT_f is the freezing point depression,
• K_f is the cryoscopic constant (freezing point depression constant) of the solvent,
• m is the molality of the solution.

First, we can rearrange the equation to find m:
m = ΔT_f / K_f
Now, we can calculate the molality m:

Given:

• ΔT_f = 0.25 K,
• K_f = 1.2 K kg mol^-1.

So:
m = 0.25 / 1.2 = 0.2083 mol/kg
Now, molality (m) is defined as:

m = (number of moles of solute) / (mass of solvent in kg)

We know:

• The mass of the solvent is 100 g = 0.1 kg,
• The mass of the solute is 5 g.

We can substitute the value of m into the formula to find the number of moles of solute:
0.2083 mol/kg = (number of moles of solute) / 0.1 kg
Number of moles of solute = 0.2083 × 0.1 = 0.02083 mol
Now, the molar mass M of the solute is given by:
M = (mass of solute) / (number of moles of solute)
Substituting the known values:
M = 5 g / 0.02083 mol = 240.0 g/mol
Thus, the molar mass of the solute is 240.0 g/mol, so the correct answer is (d) 240.0 g mol^-1.

To make the glucose solution isotonic with a 15 g/L urea solution:
Find the molarity of urea:

• Molar mass of urea = 60 g/mol.
• Molarity of urea = 15 g/L ÷ 60 g/mol = 0.25 mol/L.

Glucose molarity for isotonic solution:

• Since glucose is a non-electrolyte, its van't Hoff factor (i) is 1. To match the osmotic pressure of urea, the molarity of glucose should also be 0.25 mol/L.

Calculate mass of glucose:

• Moles of glucose needed = 0.25 mol.
• Molar mass of glucose = 180 g/mol.
• Mass of glucose = 0.25 mol × 180 g/mol = 45 g.

Thus, the mass of glucose required is 45 g.
The correct answer is (d) 45 g.

Assertion (A): Helium is used to dilute oxygen in diving apparatus is true. Helium is commonly used in helium-oxygen mixtures (like heliox) for deep-sea diving. This is because helium reduces the risk of nitrogen narcosis and allows divers to breathe more safely at greater depths.
Reason (R): Helium has a high solubility in O₂ is false. Helium has very low solubility in oxygen (and in most gases). The reason helium is used in diving mixtures is not because it has high solubility in oxygen, but because it has low solubility and doesn't cause narcosis at high pressures, unlike nitrogen.
Therefore, (A) is true, but (R) is false.

The elevation in boiling point (ΔTb) for an electrolyte solution is given by:
ΔTb = i × Kb × m
where:

• i = van't Hoff factor (number of ions produced per molecule),

• Kb = ebullioscopic constant (constant for water),

• m = molality (moles of solute per kg of solvent).

For dilute aqueous solutions (water density ≈ 1 kg/L), molarity (M) approximates molality (m). Thus, ΔTb ≈ i × Kb × M. Since Kb is constant, ΔTb depends on the product i × M.

Calculate i × M for each option:

• (a) 0.05 M NaCl: NaCl → Na⁺ + Cl^-, i = 2. i × M = 2 × 0.05 = 0.1.

• (b) 0.1 M KCl: KCl → K⁺ + Cl^-, i = 2. i × M = 2 × 0.1 = 0.2.

• (c) 0.1 M MgSO₄: MgSO₄ → Mg²⁺ + SO₄^2-, i = 3 (assuming complete dissociation). i × M = 3 × 0.1 = 0.3.

• (d) 1 M NaCl: NaCl → Na⁺ + Cl^-, i = 2. i × M = 2 × 1 = 2.

Comparison:
The lowest i × M value is 0.1 for 0.05 M NaCl, indicating the least elevation in boiling point.

Conclusion: The correct answer is (a) 0.05 M NaCl.

According to Henry's Law,
p = K_Hx
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.
∴ Solubility : Ar < CO₂ < CH₄ < HCHO

Osmotic pressure is given by:
π = i C R T
Since all are non-electrolytes, i = 1. At the same temperature, osmotic pressure depends only on molarity (moles per litre).

Step 1: Molar masses

• Glucose (C₆H₁₂O₆): 180 g mol^-1
• Urea (CH₄N₂O): 60 g mol^-1
• Sucrose (C₁₂H₂₂O₁₁): 342 g mol^-1

Step 2: Number of moles in 10 g

• Glucose: 10/180 = 0.0556 mol
• Urea: 10/60 = 0.1667 mol
• Sucrose: 10/342 = 0.0292 mol

Step 3: Molarity (in 0.250 L solution)

• P₁ (Glucose): 0.0556 / 0.250 = 0.222 M
• P₂ (Urea): 0.1667 / 0.250 = 0.667 M
• P₃ (Sucrose): 0.0292 / 0.250 = 0.117 M

Step 4: Order of osmotic pressures
P₂ (0.667) > P₁ (0.222) > P₃ (0.117)

Given, So,

Total vapour pressure of solution,

Raoult’s law:

• If the interactions between unlike molecules (A–B) are weaker than those between like molecules (A–A or B–B), the escaping tendency of molecules increases → vapour pressure of the solution is higher than ideal → positive deviation.
• If A–B interactions are stronger, vapour pressure decreases → negative deviation.

Now, analyze each option:

(a) Acetone + Chloroform: Shows negative deviation, because strong hydrogen bonding forms between C=O of acetone and H of chloroform.
(b) Chloroethane + Bromoethane: Nearly ideal behaviour, similar polarities and no strong interaction difference.
(c) Ethanol + Acetone: Shows positive deviation, because acetone breaks the H-bonding network of ethanol, reducing A–B attractions compared to A–A.
(d) Benzene + Toluene: Both are non-polar hydrocarbons with very similar forces, so nearly ideal behaviour.

ΔT_f = ik_fm
⇒ ΔTf = 1 × 5.12 × 0.078
ΔT_f = 0.3993
ΔT_f = 0.40 K

For ideal solution,
Δ_mix H = 0
Δ_mix S > 0
Δ_mix G < 0
Δ_mix V = 0

Maximum boiling azeotrope is shown by solution which shows negative deviation from Raoult's law. thanol + water shows positive deviation (minimum-boiling), while water + nitric acid shows negative deviation (maximum-boiling).

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.

_{For ideal solution, we have}
ΔH_mix = 0, ΔV_mix = 0
Now U_mix = ΔH_mix – PΔV_mix
∴ ΔU_mix = 0
_{Also, for an ideal solution,
pA = xApA^o, pB = xBpB^o
∴ Δp = pobserved – pcalculated = 0
ΔG}mix ₌ ΔH – TΔS_mix
For an ideal solumixtion, ΔSmix ≠ 0
∴ ΔG_mix ≠ 0 

_{Ba(OH)2 is a strong electrolyte and undergoes cent percent dissociation in a dilute aqueous solution.
Ba(OH)2(aq) → Ba²⁺(aq) + 2OH^–(aq)
Thus, van’t Hoff factor i = 3.} 

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^oC
p^o = 760 mm of Hg 

⇒
⇒ n₂ = 0.2046 mol
ΔT_b = K_b × m 

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 = 1/2 and x_Toluene = 1/2
p_Benzene = 1/2p^o_Benzene = 1/2 × 12.8 = 6.4 kPa
p_Toluene = 1/2p^o_Toluene = 1/2 × 3.85 = 1.925 kPa
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. 

For an ideal solution, ΔS_mix > 0 while ΔH_mix, ΔV_mix and ΔP all are 0. 

ΔT_b = iK_bm
Given, (ΔT_b)_x > (ΔT_b)_y
∴ i_xK_bm > i_yK_bm
⇒ i_x > i_y
(K_b is same for same solvent)
So, x is undergoing dissociation in water.

Ai₂(SO₄)₃ ⇌ 2Al⁺³ + 3SO₄^2–
van't Hoff factor, i = 5
K₂SO₄ ⇌ 2K⁺ + SO₄^2–
van't Hoff factor, i = 3
K₃[Fe(CN)₆] ⇌ 3K⁺ + [Fe(CN)₆]^2-
van't Hoff factor, i = 4
Al(NO₃)₃ ⇌ Al³⁺ + 3NO₃^–
van't Hoff factor, i = 4
K₄[Fe(CN)₆] ⇌ 4K⁺ + [Fe(CN)₆]^4–
van't Hoff factor, i = 5 

We know that depression in freezing point (ΔT_f ) is given as
ΔT_f = iK_fm
So, ΔT_f ∝ i
Thus, more value of i (Van’t Hoff factor), more will be depression in freezing point.
Al₂(SO₄)₃ ⇌ 2Al⁺³ + 3SO₄^2–
i is maximum i.e., 5 for Al₂(SO₄)₃

Molarity (M)

wt. of 70% acid

An ideal solution is that solution in which each component obeys Raoult’s law under all conditions of temperatures and concentrations. For an ideal solution.

The correct answer is option C
Molar mass of CH₂​Cl₂​ =12×1+1×2+35.5×2=85 g mol^–1
Molar mass of CHCl₃ ​  =12×1+1×1+35.5×3=119.5 g mol⁻¹
Moles of CH_2​Cl₂​= 40 g /85 g mol−1 = 0.47 mol
Moles of CHCl₃​ = 25.5 g /119.5 g mol−1 = 0.213 mol
Total number of moles = 0.47+0.213=0.683 mol
Mole fraction of component 2
 =0.47mol/0.683mol= 0.688
Mole fraction of component 1
=1.00–0.688=0.312 
We know that:
P_T​=p₁⁰​+(p₂⁰​−p₁⁰​)x2​
=200+(415–200)×0.688
=200+147.9
=347.9 mm Hg