Find the percentage of P in Triple super phosphate.

55.4%

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26.5%

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61.2%

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73.5%

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Solution

The correct option is B 26.5%
Molecular of weight of Triple super phosphate of lime
$= 40 + 2 (2 \times 1 + 31 + (16 \times 4)]$
$= 40 + 194$
$= 234$
% of P in triple super phosphate of lime
$= \frac{62}{234} \times 100$
$= 26.5$ %

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