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, = 127, Ag = 108, N = 14, O = 16 g )

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Solution

Q.
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 ${mol}^{- 1}$ ) [2025]

Ans.
(1)

$Molar mass of silver iodide (M_{AgI}) = (108 + 127) g/mol = 235 g/mol$

$Moles of AgI (n_{AgI}) = \frac{W_{AgI}}{M_{AgI}} = \frac{4.74}{235} mol$

$NaI(aq) + {AgNO}_{3} (aq)(excess) \to AgI(s) + {NaNO}_{3} (aq)$

As per reaction stoichiometry:

$Moles of NaI (n_{NaI}) = Moles of AgI (n_{AgI}) = \frac{4.74}{235} mol$

$Volume of NaI solution (V_{NaI}) = 20 mL = 0.02 L$

$Molarity of NaI (M_{NaI}) = \frac{n_{NaI}}{V_{NaI} (in L)} = \frac{4.74}{235 \times 0.02} mol$

$= 1.008 M$

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