JEE Mains · Chemistry · STD 11 - 6.1. Equilibrium - 1 (chemical Equilibrium)
The number of moles of \(\mathrm{NH}_{3}\), that must be added to \(2 \,\mathrm{~L}\) of \(0.80\, \mathrm{M}\, \mathrm{AgNO}_{3}\) in order to reduce the concentration of \(\mathrm{Ag}^{+}\)ions to \(5.0 \times 10^{-8}\, \mathrm{M}\left(\mathrm{K}_{\text {formation }}\right.\) for \(\left.\left[\mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}\right]^{+}=1.0 \times 10^{8}\right)\) is ...... . (Nearest integer) [Assume no volume change on adding \(\mathrm{NH}_{3}\) ]
- A \(16\)
- B \(5\)
- C \(4\)
- D \(2\)
Answer & Solution
Correct Answer
(C) \(4\)
Step-by-step Solution
Detailed explanation
Let moles added \(=\mathrm{a}\) \(\mathrm{Ag}_{(\mathrm{ag} .)}^{+}+2 \mathrm{NH}_{3(\mathrm{ag} .)} \rightleftharpoons \mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2(\mathrm{aq} .)}^{+}\) \(t=0 \quad \quad 0.8 \quad\quad\quad \left(\frac{a}{2}\right)\)…
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