Two students performed the same experiment separately, and each of them recorded two readings of mass, which are given below. The correct reading of mass is 3.0 g. On the basis of the given data, mark the correct option out of the following statements.

 Students Readings (i) (ii) A 3.01 2.99 B 3.05 2.95

 1 The results of both the students' are neither accurate nor precise. 2 The results of student A are both precise and accurate. 3 Student B's results are neither precise nor accurate. 4 The results of student B are both precise and accurate.

Subtopic:  Introduction |
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A reading on Fahrenheit scale is $200°\mathrm{F}$. The same reading on celsius scale will be :

 1 40°C 2 94°C 3 93°C 4 30°C
Subtopic:  Introduction |
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The molarity of a solution containing 5.85 g of NaCl(s) per 500 mL of solution is -

1. 4 mol L-1

2. 20 mol L-1

3. 0.2 mol L-1

4. 2 mol L-1

Subtopic:  Concentration Based Problem |
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If 500 mL of a 5 M solution is diluted to 1500 mL, the molarity of the resultant solution is:

1. 1.5 M

2. 1.66 M

3. 0.017 M

4. 1.59 M

Subtopic:  Concentration Based Problem |
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Mark the species among the following that contains the highest number of atoms.

1. 4 g He

2. 46 g Na

3. 0.40 g Ca

4. 12 g He

Subtopic:  Moles, Atoms & Electrons |
84%
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If the concentration of glucose  (C6H12O6)  in the blood is 0.9 g L-1, then the molarity of glucose in the blood is
1. 5 M
2. 50 M
3. 0.005 M
4. 0.05 M

Subtopic:  Concentration Based Problem |
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The molality of a solution containing 18.25 g of HCl gas in 500 g of water is -

 1 0.1 m 2 1 M 3 0.5 m 4 1 m
Subtopic:  Concentration Based Problem |
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The number of molecules of H2SO4 present in 100 mL of 0.02 M H2SO4 solution is:

1. $12.044×{10}^{20}$ molecules

2. $6.022×{10}^{23}$ molecules

3. $1×{10}^{23}$ molecules

4. $12.044×{10}^{23}$ molecules

Subtopic:  Concentration Based Problem |
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The mass percent of carbon in carbon dioxide is:

 1 0.034% 2 27.27% 3 3.4% 4 28.7%
Subtopic:  Concentration Based Problem |
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The empirical formula and molecular mass of a compound are ${\mathrm{CH}}_{2}\mathrm{O}$ and 180 g, respectively. The molecular formula of the compound is -

 1.  C9H18O9 2.  CH2O 3.  C6H12O6 4.  C2H4O2
Subtopic:  Empirical & Molecular Formula |
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