Question

Which among the following has highest number of
electrons?
(1) 5 g of CO2 (2) 4 g of Co
(3) 1 g of H,
(4) 6 g of Oz​

Answer 1

Answer : The highest number of electrons present in (4) 6 grams of $O_2$ has the highest number of electrons.

Explanation :

• Solution for part (1) :

Mass of $CO_2$ = 5 g

Molar mass of $CO_2$ = 44 g/mole

First we have to calculate the moles of $CO_2$.

$\text{ Moles of }CO_2=\frac{\text{ Given mass of }CO_2}{\text{ Molar mass of }CO_2}=\frac{5g}{44g/mole}=0.114moles$

Now we have to calculate the number of electrons in $CO_2$.

The number of electrons in $CO_2$ = 6 + 2(8) = 22 electrons

As, 1 mole of $CO_2$ contains $22\times (6.022\times 10^{23})$ number of electrons

So, 0.114 mole of $CO_2$ ion contains $0.114\times 22\times (6.022\times 10^{23})=15.103\times 10^{23}$ number of electrons

• Solution for part (2) :

Mass of $CO$ = 4 g

Molar mass of $CO$ = 28 g/mole

First we have to calculate the moles of $CO_2$.

$\text{ Moles of }CO=\frac{\text{ Given mass of }CO}{\text{ Molar mass of }CO}=\frac{4g}{28g/mole}=0.143moles$

Now we have to calculate the number of electrons in $CO$.

The number of electrons in $CO$ = 6 + 8 = 14 electrons

As, 1 mole of $CO$ contains $14\times (6.022\times 10^{23})$ number of electrons

So, 0.143 mole of $CO$ ion contains $0.143\times 14\times (6.022\times 10^{23})=12.056\times 10^{23}$ number of electrons

• Solution for part (3) :

Mass of $H$ = 1 g

Molar mass of $H$ = 1 g/mole

First we have to calculate the moles of $H$.

$\text{ Moles of }H=\frac{\text{ Given mass of }H}{\text{ Molar mass of }H}=\frac{1g}{1g/mole}=1moles$

Now we have to calculate the number of electrons in $H$.

The number of electrons in $H$ = 1 electron

1 mole of $H$ contains $(6.022\times 10^{23})$ number of electrons

• Solution for part (4) :

Mass of $O_2$ = 6 g

Molar mass of $O_2$ = 32 g/mole

First we have to calculate the moles of $O_2$.

$\text{ Moles of }O_2=\frac{\text{ Given mass of }O_2}{\text{ Molar mass of }O_2}=\frac{6g}{32g/mole}=0.187moles$

Now we have to calculate the number of electrons in $O_2$.

The number of electrons in $O_2$ = 2(8) = 16 electrons

As, 1 mole of $O_2$ contains $16\times (6.022\times 10^{23})$ number of electrons

So, 0.187 mole of $O_2$ ion contains $0.187\times 16\times (6.022\times 10^{23})=18.018\times 10^{23}$ number of electrons

Hence, from this we conclude that 6 grams of $O_2$ has the highest number of electrons.

Answer 2

Answer: The correct answer is Option 4.

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$      .....(1)

According to mole concept:

1 mole of an element or compound contains $6.022\time 10^{23}$  number of particles that is electrons or protons or neutrons.      ......(2)

• For 1:  5 g of $CO_2$

Given mass of carbon dioxide = 5 g

Molar mass of carbon dioxide = 44 g/mol

Putting values in equation 1, we get:

$\text{Number of moles}=\frac{5g}{44g/mol}=0.114mol$

Now, using expression 2, we get:

Number of electrons in 1 mole of carbon dioxide = [6 + (2 × 8)] = 22

0.114 moles of carbon dioxide will contain $22\times 0.114\times 6.022\times 10^{23}=15.103\times 10^{23}$ number of electrons.

Number of electrons = $15.103\times 10^{23}$

• For 2:  4 g of CO

Given mass of carbon monoxide = 4 g

Molar mass of carbon monoxide = 28 g/mol

Putting values in equation 1, we get:

$\text{Number of moles}=\frac{4g}{28g/mol}=0.143mol$

Now, using expression 2, we get:

Number of electrons in 1 mole of carbon monoxide = [6 + 8] = 14

0.143 moles of carbon monoxide will contain $14\times 0.143\times 6.022\times 10^{23}=12.0554\times 10^{23}$ number of electrons.

Number of electrons = $12.0554\times 10^{23}$

• For 3:  1 g of $H_2$

Given mass of hydrogen = 1 g

Molar mass of hydrogen = 2 g/mol

Putting values in equation 1, we get:

$\text{Number of moles}=\frac{1g}{2g/mol}=0.5mol$

Now, using expression 2, we get:

Number of electrons in 1 mole of hydrogen molecule = [2 × 1] = 2

0.5 moles of hydrogen molecule will contain $2\times 0.5\times 6.022\times 10^{23}=6.022\times 10^{23}$ number of electrons.

Number of electrons = $6.022\times 10^{23}$

• For 4:  6 g of $O_2$

Given mass of oxygen = 6 g

Molar mass of oxygen = 32 g/mol

Putting values in equation 1, we get:

$\text{Number of moles}=\frac{6g}{32g/mol}=0.1875mol$

Now, using expression 2, we get:

Number of electrons in 1 mole of oxygen = [2 × 8] = 16

0.1875 moles of oxygen will contain $8\times 0.1875\times 6.022\times 10^{23}=18.08\times 10^{23}$ number of electrons.

Number of electrons = $18.08\times 10^{23}$

Hence, the correct answer is Option 4.