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Question
Calculate the percentage composition of oxygen in lead nitrate [Pb(NO3)2]. [Pb = 207, N= 14, O = 16]
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Solution
Pb + (N)2 + (O)6
207 + 2 x 14 + 6 x 16 = 331.
So, the molecular mass of Pb(NO3)2 = 331.
331 by weight of Pb(NO3)2 contain 96 parts by weight of oxygen .
100 parts will contain = 96 x 100 / 331 = 29 ‰
So, the percentage composition of oxygen in lead nitrate is 29‰
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