ICSE Class 10 Chemistry 2023 Solved Paper

Calculate:

Question : 74 of 88

Marks: +1, -0

The percentage of phosphorus in the fertilizer super phosphate

{\Ca} ( {\H}_2 {\PO}_4) _2

correct to 1 decimal point. [At. Wt.

H=1

P=31, O=16, C a=40]

Solution:

Molecular mass of

{\Ca} ( {\H}_2 {\PO}_4) _2

\;=40+(1 × 2+31+16 × 4) × 2

\;=234 {\amu} .

Percentage of Phosphorous

\;=\;{\;\text " Mass of phosphorous in one molecule "\;}/{\;\text " Molecular mass of compound "\;}

\;=\;{62}/{234} × 100=26.49 \%

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