Question

The total number of electrons present in 8.0 g of methane is
(1) 4.8 X 10^24
(2) 3.01 X 10^24
(3) 4.8 x 10^25
(4) 3.01 * 10^23​

Answer 1

Answer:

Atomic mass of C is 12u and of H is 1u

There are 4 H atoms and 1 C atom.

Thus the molecular mass of CH4 is 12+1∗4=16u=16g/mol12+1∗4=16u=16g/mol

Now, 1 C atom contains 6 e−e−

and 1 H contains 1 e−e−

Total no. of electrons in 1 molecule of CH4=6+4=10

No. of moles in 8g CH4 is 8/16=1/2

In one mole CH4 there are 6.023*10^23

molecules.

Each molecule contains 10 electrons.

Therefore no. of electrons in in 1/2 moles of CH4=1/2*10*6.023*10^23 electrons

Therefore there are 3.01*10^23electrons in 8g CH4

Thank you

Answer 2

Answer:

The total number of electrons present in 8g of methane is equal to 30.1×10²³ electrons.

Explanation:

We know that the atomic mass of carbon = 12gmol

The atomic mass of hydrogen = 1

The molecular mass of methane = 12 +4(1)= 10=6gmol

Number of electrons in 1 molecule of CH₄ = 6+4 =10 electrons

Given, the mass of methane =8g

The number of moles in 8g  of CH₄ $=\frac{8}{16}=0.5mol$

1 mole of methane has molecules $= 6.023 *10^{23}$

0.5mol of methane has molecules $= 0.5*6.023 *10^{23}=3.0115*10^{23}$

1 molecule of methane has electrons =10

3.01×10²³ molecules of CH₄ will have electrons  $=10*3.0115*10^{23}$

                                                                             $=30.115 *10^{23}$ electrons.      

Therefore, number of electrons in 8g of methane is equal to 30.1×10²³.