MCQ
Which is the correct option for $0.1\,M,\, 500\,ml$ of $AgCl$ ?
- A$0.05\,mole$ of $AgCl$
- B$0.1\,mole$ of total ions
- C$0.05\,N_A$ number of $Cl^-$ ions
- ✓All of the above
✓
Answer
Correct option: D.
All of the above
d
${{\text{n}}_{{\text{AgCl}}}} = {\text{M}} \times {\text{V}}({\text{ litre }})$
${{\text{n}}_{{\text{AgCl}}}} = {\text{M}} \times {\text{V}}({\text{ litre }})$
${{\text{n}}_{{\text{AgCl}}}} = {\text{M}} \times {\text{V}}({\text{ litre }})$
${\text{AgCl}} \to {\text{A}}{{\text{g}}^ + } + {\text{C}}{{\text{l}}^ - }$
moles $0.05$ $0$ $0$
mole $-$ $0.05$ $0.05$
${{\text{n}}_{{\text{Total ions }}}} = 0.05 + 0.05 = 0.1$
number of $\mathrm{Cl}^{-}$ ion $=0.05 \times \mathrm{N}_{\mathrm{A}}$
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