1.

The total number of electrons in 18 mL of water ("density = 1 g "mL^(-)1) is

Answer»

`6.02xx10^(23)`
`6.02xx10^(25)`
`6.02xx10^(24)`
`6.02xx18xx10^(23)`

Solution :`"18 mL of "H_(2)O=18" g of "H_(2)O="1 mole of "H_(2)O`
`=6.02xx10^(23)" molecules"`
Electrons present in 1 MOLECULE of `H_(2)O`
`2xx1+8=10`
`therefore" Electrons present in "6.02xx10^(23)` molecules
`=10xx6.023xx10^(23)=6.02xx10^(24)`


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