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 of 1M and density 1.5 g/mL is mixed with 400 mL of water. The molarity of the H2SO4 solution will be: (final density is 1.25 g/mL )

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 |
 94%
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 |
 86%
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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A reading on Fahrenheit scale is 200°F. The same reading on celsius scale will be :
1. 40°C
2. 94°C
3. 93°C
4. 30°C

Subtopic:  Introduction |
 69%
Level 2: 60%+
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The numbers 234,000 and 6.0012 can be represented in scientific notation as:

1. 2.34×10-9and6×103

2. 0.234 ×10-6 and 60012×10-9

3. 2.34×10-9and 6.0012×10-9

4. 2.34×105 and 6.0012×100

Subtopic:  Introduction |
 85%
Level 1: 80%+
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ln the final answer of the expression \(\large{\frac{(29.2-20.2)\times(1.79\times10^5)}{1.37}}\)
the number of significant figures after solving the expression is:

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