JEE Mains · Chemistry · STD 11 - 8.3. Organic chemistry purification and characterization
By usual analysis, 1.00g of compound (X) gave 1.79g of magnesium pyrophosphate. The percentage of phosphorus in compound (X) is: (nearest integer)
(Given, molar mass in \(g~mol^{-1}\); O=16, Mg=24, P=31)
- A 50
- B 30
- C 20
- D 40
Answer & Solution
Correct Answer
(A) 50
Step-by-step Solution
Detailed explanation
% of \(P = \frac{n_{Mg_{2}P_{2}O_{7}} \times 2 \times 31}{w_{(unknown~compound)}} \times 100\) \(= \frac{(\frac{1.79}{222} \times 2 \times 31)}{1} \times 100\) \(= 49.99\% \approx 50\%\)
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