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The mass of a mole of electrons is: 0.00...

The mass of a mole of electrons is: 0.008 g 0.184 g 0.55 mg 1.673

A

0.008 g

B

0.184 g

C

0.55 mg

D

1.673

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The correct Answer is:
To find the mass of a mole of electrons, we can follow these steps: ### Step 1: Determine the mass of a single electron. The mass of an electron is given as: \[ m_e = 9.1 \times 10^{-31} \text{ kg} \] or \[ m_e = 9.1 \times 10^{-28} \text{ g} \] ### Step 2: Use Avogadro's number to find the mass of one mole of electrons. Avogadro's number (\(N_A\)) is: \[ N_A = 6.022 \times 10^{23} \text{ electrons/mole} \] ### Step 3: Calculate the mass of one mole of electrons. The mass of one mole of electrons can be calculated using the formula: \[ \text{Mass of 1 mole of electrons} = N_A \times m_e \] Substituting the values: \[ \text{Mass of 1 mole of electrons} = 6.022 \times 10^{23} \times 9.1 \times 10^{-28} \text{ g} \] ### Step 4: Perform the multiplication. Calculating the above expression: \[ \text{Mass of 1 mole of electrons} = 6.022 \times 9.1 \times 10^{-5} \text{ g} \] Calculating \(6.022 \times 9.1\): \[ 6.022 \times 9.1 = 54.8 \] Thus: \[ \text{Mass of 1 mole of electrons} = 54.8 \times 10^{-5} \text{ g} \] ### Step 5: Convert grams to milligrams. Since \(1 \text{ mg} = 10^{-3} \text{ g}\), we convert grams to milligrams: \[ 54.8 \times 10^{-5} \text{ g} = \frac{54.8 \times 10^{-5}}{10^{-3}} \text{ mg} \] This simplifies to: \[ 54.8 \times 10^{-2} \text{ mg} = 0.548 \text{ mg} \] ### Step 6: Round the answer. Rounding \(0.548 \text{ mg}\) gives us: \[ \text{Mass of 1 mole of electrons} \approx 0.55 \text{ mg} \] ### Final Answer: The mass of a mole of electrons is **0.55 mg**. ---

To find the mass of a mole of electrons, we can follow these steps: ### Step 1: Determine the mass of a single electron. The mass of an electron is given as: \[ m_e = 9.1 \times 10^{-31} \text{ kg} \] or ...
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ICSE-SOME BASIC CONCEPTS OF CHEMISTRY -OBJECTIVE (MULTIPLE CHOICE) TYPE QUESTIONS (CHOOSE THE CORRECT OPTION IN THE FOLLOWING QUESTIONS)
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  2. The number of grams of H2SO4 required to dissolve 5 g of CaCO(3) is:

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  3. The mass of a mole of electrons is: 0.008 g 0.184 g 0.55 mg 1.673

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  4. The number of atoms in 0.004 g of magnesium is close to

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  6. Chemical equation is balanced according to the law of multiple proport...

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  7. The number of water molecules in one litre of water is:

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  8. The total number of electrons present in 18 mL of water (density of wa...

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  9. 18 g of water contain:

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  10. An oxide of metal M has 40% by mass of oxygen. Metal M has relative at...

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  11. The mass of an atom of oxygen is:

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  12. 1 mole of methane (CH4) contains

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  17. A sample of Na2CO3H2O weighing 0.62 g is added to 100 mL of 0.1 N H2SO...

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  18. 2.76 g of Ag2CO3 on being heated yields a residue weighing

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  19. The number of moles of KMnO4 that are needed to react completely with ...

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  20. Normality of a '30 volume H2O2' solution is:

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