1.

How many H-atoms are present in 0.046g ofethanol

Answer»

`6 XX 10^(20)`
`1.2 xx 10^(21)`
`3 xx 10^(21)`
`3.6 xx 10^(21)`

Solution :Mol. wt of `C_(2)H_(5)OH`
`= 2 xx 12 + 5 + 16 + 1 = 46`
`:' 46g C_(2)H_(5)OH` has H atom = `6xx N_(A)`
`:' 0.046 g` of `C_(2)H_(5)OH` has `H` atoms
`= (6 xx 6.02 xx 10^(23) xx 0.046)/(46) = 3.6 xx 10^(21)`


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