Subtopic: Concentration Based Problem |
82%
Level 1: 80%+
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The significant figures present in the answer of the following calculations
is-
\(\frac{2.5 \times 1.25 \times 3.5}{2.01}\)
1. 3
2. 1
3. 2
4. 0
is-
\(\frac{2.5 \times 1.25 \times 3.5}{2.01}\)
1. 3
2. 1
3. 2
4. 0
Subtopic: Introduction |
68%
Level 2: 60%+
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The average atomic mass of hydrogen using the following data is:
1. 1.00015 u
2. 1.015 u
3. 1.0015 u
4. 1.0005 u
| Isotope | % Natural Abundance | Molar mass |
| 1H | 99.985 | 1 |
| 2H | 0.015 | 2 |
1. 1.00015 u
2. 1.015 u
3. 1.0015 u
4. 1.0005 u
Subtopic: Introduction |
61%
Level 2: 60%+
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Calcium carbonate reacts with aqueous HCl to give CaCl2 and CO2 according to the reaction given below:
CaCO3 (s) + 2HCl (aq) → CaCl2(aq) + CO2(g) + H2O(l)
The mass of CaCl2 formed when 250 mL of 0.76 M HCl reacts with 1000 g of CaCO3 will be -
1. 2.06 g
2. 20.6 g
3. 1.54 g
4. 10.54 g
CaCO3 (s) + 2HCl (aq) → CaCl2(aq) + CO2(g) + H2O(l)
The mass of CaCl2 formed when 250 mL of 0.76 M HCl reacts with 1000 g of CaCO3 will be -
1. 2.06 g
2. 20.6 g
3. 1.54 g
4. 10.54 g
Subtopic: Limiting Reagent |
62%
Level 2: 60%+
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The pair of molecules that does not endorse law of multiple proportions is -
1. CO and CO2
2. H2O2 and H2O
3. SO2 and SO3
4. NO2- ion and NO2
1. CO and CO2
2. H2O2 and H2O
3. SO2 and SO3
4. NO2- ion and NO2
Subtopic: Introduction |
77%
Level 2: 60%+
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Subtopic: Concentration Based Problem |
87%
Level 1: 80%+
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Calculate the mass of 95% pure CaCO3 that will be required to neutralize 50 mL of 0.5 M HCl solution according to the following reaction.
CaCO3(s) + 2HCl(aq) → CaCl2(aq) + CO2(g) + 2H2O(l)
[Calculate up to the second place of decimal point]
CaCO3(s) + 2HCl(aq) → CaCl2(aq) + CO2(g) + 2H2O(l)
[Calculate up to the second place of decimal point]
| 1. | 9.50 g | 2. | 1.25 g |
| 3. | 1.32 g | 4. | 3.65 g |
Subtopic: Equation Based Problem |
55%
Level 3: 35%-60%
NEET - 2022
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Match List-I with List-II:
| List-I (Quantities) |
List-II (Corresponding Values) |
||
| (a) | 4.48 litres of O2 at STP | (i) | 0.2 mole |
| (b) | 12.022 × 1022 molecules of H2O | (ii) | 12.044 × 1023 molecules |
| (c) | 96 g of O2 | (iii) | 6.4 g |
| (d) | 88 g of CO2 | (iv) | 67.2 litres at STP |
Choose the correct answer from the options given below:
| (a) | (b) | (c) | (d) | |
| 1. | (i) | (iii) | (iv) | (ii) |
| 2. | (iii) | (i) | (iv) | (ii) |
| 3. | (iv) | (i) | (ii) | (iii) |
| 4. | (iii) | (i) | (ii) | (iv) |
Subtopic: Moles, Atoms & Electrons |
50%
Level 3: 35%-60%
NEET - 2022
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Assertion: Specific gravity is dimensionless.
Reason: Specific gravity is the density of a substance measured with respect to density of water at 4°C.
1. Both statements are true and the reason is the correct explanation of the assertion.
2. Both statements are true but the reason is not the correct explanation of the assertion.
3. Assertion is true and the reason is false.
4. Assertion is false and reason is true.
Reason: Specific gravity is the density of a substance measured with respect to density of water at 4°C.
1. Both statements are true and the reason is the correct explanation of the assertion.
2. Both statements are true but the reason is not the correct explanation of the assertion.
3. Assertion is true and the reason is false.
4. Assertion is false and reason is true.
Subtopic: Introduction |
71%
Level 2: 60%+
Hints
| Assertion (A): | The molarity of pure water is 55.55 M at 298 K. |
| Reason (R): | Molarity is temperature-dependent. |
| 1. | Both (A) and (R) are True and (R) is the correct explanation of (A). |
| 2. | Both (A) and (R) are True but (R) is not the correct explanation of (A). |
| 3. | (A) is True but (R) is False. |
| 4. | (A) is False but (R) is True. |
Subtopic: Concentration Based Problem |
62%
Level 2: 60%+
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