The percentage of phosphorous in Ca3(PO4)2 is (Ca=40, P=31, O=16)
(i) 25%
(ii) 20%
(iii) 23.5%
(iv) 20.5%
Answers
Answered by
11
molecular weight of compound = 120 + 62 + 128 = 310
weight of phosphurus in compound = 62
% of phosphurus in compound = weight of phosphurus ×100/molecular weight
= 62×100/310 = 20%
Answered by
0
Answer:
Ca3(PO4)2
=40x3+(31+64)x2
=120+(95)x2
=120+190
=310g/mol
Atomic mass of phosphorus (p)=31g
% of phosphorus=2x31/310x100
=20%
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