Question

1.8 ग्राम जल में इलेक्ट्रॉनों की संख्या कितनी होगी?

Answer 1

$6.023 \times 10^{23}$ Electrons

Explanation:

$Number\ of\ moles = \frac {Given\ Mass}{Molar\ Mass}$

Therefore, 1.8 ml of water is 1.8 gram of water so, we using the formula above

No. of moles of water = $\frac {1.8}{18} =0.1$

Therefore, 1 atom of water have 2+8 = 10 electron

So, Number of atoms of water = $no.of moles \times av. Constant$

= $0.1 \times 6.022 \times10^{23} = 0.6022 \times 10^{23}$ atoms of water

• Therefore, 1 atoms of water is 10 electron contain

So, $0.6022 \times 10^{23}$ atoms of water contain = $0.6022 \times 10^{23} \times 10 = 6.02 \times 10^{23}$ electrons

Answer 2

Structure of atoms

Number of electrons in 1.8 gm of water.

As we know that molecular formula for water can be written as $H_2O$.

1 molecule of water contains 2 hydrogen and 1 oxygen atoms,

H atoms has 1 electron, O atoms has 8 electrons .

Hence $H_2O$ molecule has $10 \ e^-$.

1 mole of water molecule $=$ 10 mole electrons.

$18 \ gm \ H_2O = 1 \ mole\ H_2O$

$Number\ of\ moles=\frac{given\ weight}{molecular\ weight}$,

given weight is 1.8 gm.

molecular weight is 18 gm.

$number\ of\ moles=\frac{1.8}{18}=0.1\ mole$

Therefore Avogadro number of molecules in 1 mole are $0.1\times6.022\times10^{23}$ $H_2O$ molecule.

Number of electron in $H_2O$ is 10 electron.

Number of electrons  in 0.1 gm $H_2O$=$0.1\times 10\times 6.023\times 10^{23}=6.023\times 10^{23} \ e^-$.