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How many moles of `O` are present in `4.9 g` of `H_(3) PO_(4)`? (Atomic weight of `P, O` and `H = 31, 16, 1`) |
Answer» Molecular weight of `H_(3) PO_(4) = 1 xx 3 + 31 + 16 xx 4 = 98 g` `98 g = 1 "mole of" H_(3) PO_(4) = 4` mole of `O` `4.9 g of H_(3) PO_(4) = (4)/(98) xx 4.9 = 0.2` mole of O |
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