JEE Mains · Chemistry · STD 11 - 1. Some basic concept of chemistry
The volume \((\text { in } \mathrm{mL})\) of \(0.125\; \mathrm{M}\; \mathrm{AgNO}_{3}\) required to quantitatively precipitate chloride ions in \(0.3\; \mathrm{g}\) of \(\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}\right] \mathrm{Cl}_{3}\) is \(^{\mathrm{M}}\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}\right] \mathrm{Cl}_{3}=267.46 \;\mathrm{g} / \mathrm{mol}\) \(\mathrm{M}_{\mathrm{AgNO}_{3}}=169.87 \;\mathrm{g} / \mathrm{mol}\)
- A \(32.06\)
- B \(38.25\)
- C \(26.92\)
- D \(24.34\)
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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