JEE Mains · Chemistry · STD 12 -1. Solution and colligative properties
20 mL of sodium iodide solution gave 4.74 g silver iodide when treated with excess of silver nitrate solution. The molarity of the sodium iodide solution is ________ M. (Nearest Integer value) (Given : \(\mathrm{Na}=23, \mathrm{I}=127, \mathrm{Ag}=108, \mathrm{~N}=14\), \(\mathrm{O}=16 \mathrm{~g} \mathrm{~mol}^{-1}\))
- A 5
- B 4
- C 3
- D 1
Answer & Solution
Correct Answer
(D) 1
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & \mathrm{NaI}_{(\mathrm{aq})}+\mathrm{AgNO}_{3(\mathrm{aq)}} \rightarrow \mathrm{AgI}_{(\mathrm{s})}+\mathrm{NaNO}_3(\mathrm{aq}) \\ & \mathrm{M}, 20 \mathrm{ml} \text { excess } \\ & 4.74 \mathrm{~g} \end{aligned}\)…
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