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ICSE Class 10 Chemistry 2023 Paper

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Question : 74 of 88
Marks: +1, -0
The percentage of phosphorus in the fertilizer super phosphate Ca(H2PO4)2\mathrm{Ca}(\mathrm{H}_2\mathrm{PO}_4)_2 correct to 1 decimal point. [At. Wt. H=1\mathrm{H}=1 , P=31,O=16,Ca=40]\mathrm{P}=31, \mathrm{O}=16, \mathrm{Ca}=40]
Solution:  
Molecular mass of Ca(H2PO4)2\mathrm{Ca}(\mathrm{H}_2\mathrm{PO}_4)_2
  =40+(1×2+31+16×4)×2\;=40+(1 \times 2+31+16 \times 4) \times 2
  =234amu.\;=234 \mathrm{amu}.
Percentage of Phosphorous
  =  Mass of phosphorous in one molecule    Molecular mass of compound  \;=\frac{\;\text{Mass of phosphorous in one molecule}\;}{\;\text{Molecular mass of compound}\;}
  =62234×100=26.49%\;=\frac{62}{234} \times 100=26.49\%
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