1.

The number of water molecules present in a drop of water weighing 0.018g is

Answer»

`6.022 xx 10^(26)`
`6.022 xx 10^(23)`
`6.022 xx 10^(19)`
`6.022 xx 10^(20)`

Solution :No. of moles = `("Moles")/("MOL. Mass") = (0.018)/(18) = 1 xx 10^(-3)`
No. of moles = `("No. of molecules")/(N_(0))`
`1 xx 10^(-3) = ("No. of molecules")/(6.022 xx 10^(23)) = 6.022 xx 10^(20)`


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