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NEET Chemistry MCQ
Quiz 1
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Q.1
Find the molarity of 5% aq. sucrose solution of density 1.1 gm/ml.
0%
.2 M
0%
.14 M
0%
.3 M
0%
.5 M
Explanation
Solution: Sucrose (C₁₂H₂₂O₁₁) has molar mass 342 g/mol. A 5% solution means 5 g sucrose per 100 mL of solution. Moles of sucrose = 5/342 = 0.0146 mol Molarity = moles / volume(L) = 0.0146 / 0.100 ≈ 0.146 M ≈ 0.14 M (using the density to confirm the solution's volume is essentially 100 mL for this mass, i.e. close to that of water).
Q.2
An ideal solution is formed when its components
0%
Have no change in volume on mixing
0%
Have no change in enthalpy on mixing
0%
have the both the above characteristics
0%
Have high solubility
Explanation
An ideal solution is one where the interactions between different components are essentially the same as between like components — which means mixing produces no volume change (ΔV = 0) and no enthalpy change (ΔH = 0). Both conditions must hold together, not just one.
Q.3
Colligative properties are observed when _____________. (More than one answer)
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a non volatile solid is dissolved in a volatile liquid.
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a non volatile liquid is dissolved in another volatile liquid.correct
0%
a gas is dissolved in non volatile liquid.
0%
a volatile liquid is dissolved in another volatile liquid
Explanation
Colligative properties depend only on the NUMBER of solute particles, not their identity — which requires the solute itself to not contribute its own vapour pressure. This holds whenever a non-volatile substance — whether solid (e.g. sugar in water) or liquid (e.g. glycerol in water) — is dissolved in a volatile solvent; both scenarios genuinely produce measurable colligative effects.
Q.4
which of the following is incorrect for ideal solution
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$\Delta H =0$
0%
$\Delta G =0$
0%
$\Delta U =0$
0%
$P =P_{obs} – P _{\text{Rauolt law}}=0$
Explanation
For an ideal solution, mixing occurs with no volume change (ΔV = 0) and no enthalpy change (ΔH = 0), and consequently no internal energy change (ΔU = 0) either. It obeys Raoult's law exactly, so the observed pressure matches the Raoult's-law prediction (their difference is zero). But mixing is still spontaneous — entropy increases (ΔS > 0) — so ΔG = ΔH − TΔS = −TΔS is negative, NOT zero. So "ΔG = 0" is the incorrect statement.
Q.5
If the molality of the dilute solution is tripled, the molal depression constant will become
0%
1/3
0%
3
0%
6
0%
remain unchanged
Explanation
The molal depression constant (Kf) is an intrinsic property of the solvent alone — it doesn't depend on how much solute is dissolved. Changing the molality of the solution has no effect on Kf; it stays the same.
Q.6
The mole fraction of solute in 1 molal aqueous solution is
0%
.009
0%
.018
0%
.027
0%
.036
Explanation
1 molal means 1 mole of solute per 1000 g (1 kg) of water. Moles of water = 1000/18 = 55.56 mol Mole fraction of solute = 1 / (1 + 55.56) = 1/56.56 ≈ 0.0177 ≈ 0.018.
Q.7
We have three aqueous solutions of NaCl labelled as ‘A’, ‘B’ and ‘C’ with concentrations 0.1M, 0.01M and 0.001M, respectively. The value of van’t Hoff factor for these solutions will be in the order______.
0%
$i_A < i_B < i_C$
0%
$i_A > i_B > i_C$
0%
$i_A =i_B = i_C$
0%
$i_A < i_B > i_C$
Explanation
For a strong electrolyte like NaCl, which is essentially fully dissociated, the van't Hoff factor stays close to its ideal value (i ≈ 2) across this whole dilute concentration range, regardless of whether the solution is 0.1 M, 0.01 M or 0.001 M — so the three values are essentially equal.
Q.8
The vapour pressure of water at 293 K is 0.0231 bar & the vapour pressure of the solution of 108.24 gm of a compound in 1000 gm of water at the same temperature is 0.0228 bar, find the molar mass of the solute
0%
160 g
0%
150 g
0%
147 g
0%
142 g
Explanation
Using Raoult's law for the relative lowering of vapour pressure: (P° − P)/P° = mole fraction of solute (0.0231 − 0.0228)/0.0231 = 0.0003/0.0231 = 0.01299 Moles of water = 1000/18 = 55.56 mol. Let n = moles of solute: n/(n + 55.56) = 0.01299 → n = 0.01299(n + 55.56) → 0.987n = 0.7215 → n ≈ 0.731 mol Molar mass = mass/moles = 108.24 / 0.731 ≈ 147–148 g/mol.
Q.9
which of the following salt has the same value of Vant’s hoff factor(i) as that of $K_3[Fe(CN)_6]$
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$Na_2SO_4$
0%
HCL
0%
$AL(NO_3)_3$
0%
$AL_2(SO_4)_4$
Explanation
K₃[Fe(CN)₆] dissociates into 3 K⁺ ions + 1 [Fe(CN)₆]³⁻ ion = 4 particles total, so i = 4. Checking the options: Na₂SO₄ → 2 Na⁺ + SO₄²⁻ = 3 particles (i = 3). HCl → H⁺ + Cl⁻ = 2 particles (i = 2). Al(NO₃)₃ → Al³⁺ + 3 NO₃⁻ = 4 particles (i = 4) — this matches K₃[Fe(CN)₆] exactly.
Q.10
Match the column
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p ->ii, q -> iv, r -> iii, s->i
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p ->ii, q -> iv, r -> ii, s->i
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p ->i, q -> iv, r -> ii, s->iii
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p ->i, q -> iv, r -> ii, s->ii
Explanation
Matching each quantity to its units: (p) Ebullioscopic constant (Kb): from ΔTb = Kb × molality, units = K/(mol/kg) = K·kg/mol → (ii) (q) Henry's constant: in the form p = KH × x (x is a dimensionless mole fraction), KH carries units of pressure → (iv) Bar (r) Cryoscopic constant (Kf): same relationship/units as Kb → also K·kg/mol → (ii) (s) Gas constant R: R = 8.314 J/(mol·K) = N·m/(mol·K) → (i) So: p → ii, q → iv, r → ii, s → i (Kb and Kf genuinely share the same units, which is why both map to the same option).
Q.11
Assertion and Reason (a) Assertion and reason both are correct statements and reason is correct explanation for assertion. (b) Assertion and reason both are correct statements but reason is not correct explanation for assertion. (c) Assertion is correct statement but reason is wrong statement. (d) Assertion and reason both are incorrect statements. (e) Assertion is wrong statement but reason is correct statement. Assertion : When a solution is separated from the pure solvent by a semipermeable membrane, the solvent molecules pass through it from pure solvent side to the solution side. Reason : Diffusion of solvent occurs from a region of high concentration solution to a region of low concentration solution.
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(a)
0%
(b)
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(c)
0%
(d)
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(e)
Explanation
The Assertion correctly describes osmosis: solvent moves through a semipermeable membrane from the pure solvent side into the solution. The Reason states a general diffusion principle (movement from a region of higher concentration to lower concentration) that is true in a broad sense, but it doesn't precisely capture what actually drives osmosis — which is the difference in solvent chemical potential/effective solvent concentration across the membrane, not diffusion in the ordinary bulk sense. So both statements are true, but the Reason isn't the precise, correct explanation of the Assertion.
Q.12
Assertion and Reason (a) Assertion and reason both are correct statements and reason is correct explanation for assertion. (b) Assertion and reason both are correct statements but reason is not correct explanation for assertion. (c) Assertion is correct statement but reason is wrong statement. (d) Assertion and reason both are incorrect statements. (e) Assertion is wrong statement but reason is correct statement. Assertion : the vapour pressure of the liquid decrease if non- volatile substance is added to the liquid Reason : Relative lowering vapour pressure is equal to Mole fraction of solute
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(a)
0%
(b)
0%
(c)
0%
(d)
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(e)
Explanation
The Assertion is correct: adding a non-volatile solute does lower the solvent's vapour pressure. The Reason states the correct quantitative relationship — relative lowering of vapour pressure equals the mole fraction of solute (Raoult's law) — but this is a mathematical formula describing the SIZE of the effect, not itself an explanation of the underlying physical reason vapour pressure drops (solute molecules reducing the solvent's effective mole fraction/escaping tendency at the surface). So both are true, but the Reason doesn't directly explain the Assertion.
Q.13
Assertion and Reason (a) Assertion and reason both are correct statements and reason is correct explanation for assertion. (b) Assertion and reason both are correct statements but reason is not correct explanation for assertion. (c) Assertion is correct statement but reason is wrong statement. (d) Assertion and reason both are incorrect statements. (e) Assertion is wrong statement but reason is correct statement. Assertion: An unripe mango placed in a concentrated salt solution to prepare pickle, shrivels Reason: Mango loses water due to reverse osmosis.
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(a)
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(b)
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(c)
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(d)
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(e)
Explanation
The Assertion is true — an unripe mango placed in concentrated salt solution does shrivel, as water moves out of its cells into the surrounding hypertonic solution. But this is ordinary osmosis (specifically exosmosis, water leaving the cell), not "reverse osmosis" — reverse osmosis is a distinct, engineered process where external pressure is applied to force water AGAINST its natural osmotic direction. No such applied pressure is involved here, so the Reason's terminology is incorrect.
Q.14
Assertion and Reason (a) Assertion and reason both are correct statements and reason is correct explanation for assertion. (b) Assertion and reason both are correct statements but reason is not correct explanation for assertion. (c) Assertion is correct statement but reason is wrong statement. (d) Assertion and reason both are incorrect statements. (e) Assertion is wrong statement but reason is correct statement. Assertion : Molarity of a solution in liquid state changes with temperature. Reason : The volume of a solution changes with change in temperature
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(a)
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(b)
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(c)
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(d)
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(e)
Explanation
Molarity = moles of solute / volume of solution. The moles of solute don't change with temperature, but the solution's volume does (due to thermal expansion or contraction) — so molarity changes with temperature too. The Reason directly and correctly explains why: it's precisely the volume term in the molarity formula that shifts with temperature.
Q.15
Link Comprehension type The degree of dissociation of $ Ca(NO_3)_2$ in a dilute aqueous solution containing 7 g of the slat in 100 g of the solution at 100° C is 70 percent. The vapour pressure of the water at 100° C is 760 mm. Find the Vant Off factor
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2.4
0%
3.2
0%
1.4
0%
1.6
Explanation
For Ca(NO₃)₂, which dissociates into 3 ions (1 Ca²⁺ + 2 NO₃⁻), the van't Hoff factor with degree of dissociation α is: i = 1 + α(n − 1) = 1 + 0.70 × (3 − 1) = 1 + 1.4 = 2.4.
Q.16
Find the Vapour pressure of the solution
0%
720 mm
0%
745.3 mm
0%
754 mm
0%
650 mm
Explanation
Continuing from the previous part (i = 2.4): with 7 g Ca(NO₃)₂ (molar mass 164 g/mol) in 93 g of water (100 g solution − 7 g solute): n(solute) = 7/164 = 0.0427 mol; n(water) = 93/18 = 5.167 mol Mole fraction of solute = 0.0427/(0.0427 + 5.167) ≈ 0.00819 Relative lowering of vapour pressure = i × x(solute) = 2.4 × 0.00819 ≈ 0.01966 P° − P = 0.01966 × 760 ≈ 14.9 mm P (solution) = 760 − 14.9 ≈ 745.1 mm — matching the option 745.3 mm within rounding.
Q.17
Link Comprehension type The vapour pressures of ethanol and methanol are 44.5 mm and 88.7 mm Hg respectively. An ideal solution is formed at the same temperature by mixing 60 g of ethanol with 40 g of methanol Calculate the total vapour pressure of the solution
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46.13 mm Hg
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56.25 mm Hg
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66.13 mm Hg
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80.15 mm Hg
Explanation
Moles: ethanol (MW 46) = 60/46 = 1.304 mol; methanol (MW 32) = 40/32 = 1.25 mol. Total = 2.554 mol. Mole fractions: x(ethanol) = 1.304/2.554 ≈ 0.511; x(methanol) ≈ 0.489. By Raoult's law, partial pressures: P(ethanol) = 0.511 × 44.5 ≈ 22.73 mm Hg; P(methanol) = 0.489 × 88.7 ≈ 43.40 mm Hg. Total vapour pressure = 22.73 + 43.40 ≈ 66.13 mm Hg.
Q.18
Calculate the Mole fraction of ethanol in vapour
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.3437
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.6563
0%
.437
0%
.537
Explanation
Using Dalton's law, the mole fraction of a component IN THE VAPOUR equals its partial pressure divided by the total pressure: y(ethanol) = P(ethanol) / P(total) = 22.73 / 66.13 ≈ 0.3437.
Q.19
Find the Vapour pressure of Component A and B
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12.8mmHg, 43.40 mm hg
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10.73mmHg, 73.40 mm hg
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22.73mmHg, 53.40 mm hg
0%
22.73 mm Hg, 43.40 mm hg
Explanation
From the Raoult's-law calculation above: partial vapour pressure of ethanol (A) ≈ 22.73 mm Hg, and of methanol (B) ≈ 43.40 mm Hg.
Q.20
A solution of acetone in ethanol
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behaves like a near ideal solution
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obeys Raoults law
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shows a negative deviation from Raoults law
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shows a positive deviation from Raoults law
Explanation
A mixture of acetone and ethanol shows a positive deviation from Raoult's law — acetone-ethanol interactions are weaker than the ethanol-ethanol hydrogen bonding (and acetone-acetone interactions) present in the pure components, so molecules escape into the vapour phase more readily than an ideal mixture would predict, giving a higher total vapour pressure than the ideal (Raoult's law) value.
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