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

\(\left[ Co \left( NH _{3}\right)_{6}\right] Cl _{3}\) के \(0.3 \,g\) में क्लोराइड आयन को मात्रात्मक रूप से अवक्षेपित करने के लिए \(0.125 M AgNO _{3}\) का कितना आयतन \(( mL\) में) आवश्यक होगा ............. \(M \left[ Co \left( NH _{3}\right)_{6}\right] Cl _{3}=267.46 \,g / mol\) \({ }^{ M } AgNO _{3}=169.87 \,g / mol\)

  1. A \(32.06\)
  2. B \(38.25\)
  3. C \(26.92\)
  4. D \(24.34\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(26.92\)

Step-by-step Solution

Detailed explanation

Number of moles of \(\mathrm{Cl}^{-}\) precipitated in \(\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}\right] \mathrm{Cl}_{3}\) is equal to number of moles of \(\mathrm{AgNO}_{3}\) used. \(\frac{0.3}{267.46} \times 3=\frac{0.125 \times \mathrm{V}}{1000}\) where…
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