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Calculate the Percentage of Phosphorous in the Fertilizer Superphosphate, Ca(H2po4)2. Ca = 40, H =1, P =31, O = 16 (Correct to 1 Decimal Place)

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Question

Calculate the percentage of phosphorous in the fertilizer superphosphate, Ca(H2PO4)2. [Ca = 40, H =1, P =31, O = 16] (Correct to 1 decimal place)

Sum
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Solution

Molecular mass of fertilizer superphosphate, Ca(H2PO4)2 =234
234 parts by weight of fertilizer contains 62 parts by weight of phosphorous
So, 100 parts will contain = 62 x 100/234 = 26.5‰

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Chapter 5: Mole Concept and Stoichiometry - Exercise 5 [Page 119]

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Frank Chemistry Part 2 [English] Class 10 ICSE
Chapter 5 Mole Concept and Stoichiometry
Exercise 5 | Q 6 | Page 119

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