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JEE Mains · Chemistry · STD 11 - 1. Some basic concept of chemistry

\(C _7 H _5 N _3 O _6\) के \(681\) ग्राम में उपस्थित \(N\) परमाणुओ की संख्या \(x \times 10^{21}\) है। \(x\) का मान \(.........\) है। (निकटतम पूर्णांक) (दिया है : \(N _{ A }=6.02 \times 10^{23}\,mol ^{-1}\) )

  1. A \(6418\)
  2. B \(5418\)
  3. C \(5118\)
  4. D \(5948\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(5418\)

Step-by-step Solution

Detailed explanation

M.M. of \(C _{7} H _{5} N _{3} O _{6}\) is \(84+5+42+96=227\) \(n _{ C _{7} H _{5} N _{3} O _{6}}=\frac{681}{227}=3\) \(n _{ N }=\frac{681}{227} \times 3=9 mol\) no. of \(N\) atoms \(=9 \times 6.02 \times 10^{23}\) \(=5418 \times 10^{21}\) \(\therefore\) The answer is \(5418 .\)
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