The molality of a 15% (w/vol.) solution of H2SO4 of density 1.1g/cm3 is:

1. 1.2 mol/kg

2. 1.4 mol/kg

3. 1.8 mol/kg

4. 1.6 mol/kg

Subtopic:  Concentration Based Problem |
 51%
Level 3: 35%-60%
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100 mL of H₂SO₄ solution having molarity 1 M and density 1.5 g mL⁻¹ is mixed with 400 mL water.
If the density of the final solution is 1.25 g mL⁻¹, calculate the molarity of the final H₂SO₄ solution.

1. 4.4 M 2. 0.145 M
3. 0.52 M 4. 0.227 M
Subtopic:  Moles, Atoms & Electrons |
 57%
Level 3: 35%-60%
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Assertion (A): 1 g O2 and 1 g O3  have an equal number of oxygen atoms.
Reason (R): O2  and  O3  have different molar masses.
 
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:  Moles, Atoms & Electrons |
Level 3: 35%-60%
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Match the following physical quantities with units.

Physical quantity Unit
A. Molarity (i) mol kg–1 
B.  Molality (ii) mol L–1
C.  Pressure (iii) Candella
D. Luminous intensity (iv) Pascal

Codes:

A B C D
1. (i) (iv) (ii) (iii)
2. (ii) (i) (iv) (iii)
3. (i) (iv) (iii) (ii)
4. (iv) (i) (iii) (ii)
Subtopic:  Introduction |
 93%
Level 1: 80%+
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The correct match is:
List-I List-II
a. Micro i.  10-15 m
b. Mega ii.  10-6 m
c. Giga iii.  106 m
d. Femto iv.  109 m
Codes:
a b c d
1. i iv iii ii
2. iii iv ii i
3. ii iii iv i
4. i iii iv ii
Subtopic:  Introduction |
 85%
Level 1: 80%+
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At 100 ºC and 1 atm, if the density of liquid water is 1.0 g cm–3 and that of water vapor is 0.0006 g cm-3, then the volume occupied by water molecules in 1 litre of steam at that temperature will be:

1. 6 cm 2. 60 cm3
3. 0.6 cm3  4. 0.06 cm3
Subtopic:  Moles, Atoms & Electrons |
 60%
Level 2: 60%+
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At STP, the density of CCl4 vapour in g/L will be closest to: 

1. 8.67

2. 6.87

3. 3.67

4. 4.26

Subtopic:  Concentration Based Problem |
 76%
Level 2: 60%+
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Find the Celsius scale reading corresponding to a temperature of 200°F.
1. 40°C
2. 94°C
3. 93°C
4. 30°C

Subtopic:  Introduction |
 68%
Level 2: 60%+
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The numbers 234,000 and 6.0012 can be represented in scientific notation as:
 

1. \(2.34 \times 10^{-9} \text { and } 6 \times 10^3\)
2. \(0.234 \times 10^{-6} \text { and } 60012 \times 10^{-9}\)
3. \(2.34 \times 10^{-9} \text { and } 6.0012 \times 10^{-9}\)
4. \(2.34 \times 10^5 \text { and } 6.0012 \times 10^0\)
Subtopic:  Introduction |
 83%
Level 1: 80%+
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Determine the number of significant figures in the final answer of the expression \(\large{\frac{(29.2-20.2)\times(1.79\times10^5)}{1.37}}\):

1. 2 2. 4
3. 6 4. 7
Subtopic:  Introduction |
 60%
Level 2: 60%+
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