1.

Urea is a very important nitrogenous fertiliser. Its formula is CON_(2)H_(4). Calculate the percentage of nitrogen in urea. (C = 12, O = 16, N = 14 and H = 1)

Answer»

Solution :`CO(NH_(2))_(2)`
Molecular mass = `12+16+2(14+2)=60`
PERCENTAGE of NITROGEN = `("Mass of nitrogen in the compound")/("Molecular mass")XX100`
= `28/60xx100=46.67%`


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