JEE MAIN - Chemistry (2025 - 8th April Evening Shift - No. 24)
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 : Na = 23, I = 127, Ag = 108, N = 14, O = 16 g mol-1)
Answer
1
Explanation
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$$\begin{aligned} & \text { Moles of } \mathrm{I}^{-} \text {in } \mathrm{NaI}=\text { Moles of }\left(\mathrm{I}^{-}\right) \text {in } \mathrm{AgI}=\frac{4.74}{235} \\ & \text { Moles of } \mathrm{NaI}=\frac{4.74}{235} \\ & \text { Molarity }[\mathrm{NaI}]=\frac{4.74}{235 \times 0.02}=1.008 \end{aligned}$$
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