Question

Calculate the number of electrons present in 36 g water .

Answer 1

Molar mass of $H_{2}O$ = 18 g

No. of molecules present in 1 mole of $H_{2}O$ = $6.022\times {10}^{23}$

No. of electrons present in 1 molecule of $H_{2}O$ = 10

No. of moles of $H_{2}O$ = $\frac{36}{18}=2$

So, the no. of molecules of $H_{2}O$ present in 2 moles of $H_{2}O$ = $2\times 6.022\times {10}^{23}$

No. of electrons present in 36 g of water = $10\times 2\times 6.022\times {10}^{23}=1.2044\times {10}^{25}$