How many atoms of each type present in 143 g of washing soda `(Na_(2)CO_(3).10H_(2)O)` ?

Explanation

`n = (w)/(M_(0)) = (143)/(286) = 0.5` mole, Na atoms `= 0.5 xx 2 xx N_(A) = 6.022 xx 10^(23)`, H atoms `= 0.5 xx 20 xx N_(A) = 10^(24)`
C taoms `= 0.5 N_(A) = 3.011 xx 10^(23)` , O atoms `= 0.5 xx 13 xx N_(A) = 3.91 xx 10^(24)`