ExamBro
ExamBro
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)

  1. A 50
  2. B 30
  3. C 20
  4. D 40
Verified Solution

Answer & Solution

Correct Answer

(A) 50

Step-by-step Solution

Detailed explanation

% of \(P = \frac{n_{Mg_{2}P_{2}O_{7}} \times 2 \times 31}{w_{(unknown~compound)}} \times 100\) \(= \frac{(\frac{1.79}{222} \times 2 \times 31)}{1} \times 100\) \(= 49.99\% \approx 50\%\)
Same subject
Explore more questions on app
From JEE Mains
Explore more questions on app