20 mL of sodium iodide solution gave 4.74 g | vedclass

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(C) The chemical reaction is: $NaI_{(aq)} + AgNO_{3(aq)} \rightarrow AgI_{(s)} + NaNO_{3(aq)}$$1$. Calculate the molar mass of $AgI$: $108 + 127 = 235

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$1$

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(C) The chemical reaction is: $NaI_{(aq)} + AgNO_{3(aq)} \rightarrow AgI_{(s)} + NaNO_{3(aq)}$$1$. Calculate the molar mass of $AgI$: $108 + 127 = 235 \ g \ mol^{-1}$.$2$. Calculate the moles of $AgI$ formed: $n(AgI) = \frac{4.74 \ g}{235 \ g \ mol^{-1}} \approx 0.02017 \ mol$.$3$. From the stoichiometry,$1 \ mol$ of $NaI$ produces $1 \ mol$ of $AgI$. Therefore,moles of $NaI = 0.02017 \ mol$.$4$. Calculate the molarity $(M)$: $M = \frac{n}{V(L)} = \frac{0.02017 \ mol}{0.020 \ L} = 1.0085 \ M$.$5$. The nearest integer value is $1$.

$20 \ mL$ 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 $........ M$. $(Nearest \ Integer \ value)$
$(Given: Na=23, I=127, Ag=108, N=14, O=16 \ g \ mol^{-1})$

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$20 \ mL$ 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})$