An organic compound 'A' is oxidized with Na_{2}O_{2} followed by boiling with HNO_{3}. The resultant solution is then treated with ammonium molybdate to yield a yellow precipitate.

Based on above observation, the element present in the given compound is : 


 

  • Option 1)

    Sulphur

  • Option 2)

    Nitrogen

  • Option 3)

    Phosphorus 

  • Option 4)

    Fluorine

Answers (1)

According to question : 

A+Na_{2}O_{2}\rightarrow y\left ( Resultant\: \: solution \right )

Y+HNO_{3}\overset{\left ( NH_{4} \right )_{2}M_{o}0_{4}}{\rightarrow}yellow\: \: ppt.

\left ( NH_{4} \right )_{2}MoO_{4} is used for detection of phosphorus. This yellow coloured precipitate helps us to conclude that Phosphorus is present in the organic compound. 


Option 1)

Sulphur

Option 2)

Nitrogen

Option 3)

Phosphorus 

Option 4)

Fluorine

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