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$^{-1}$)
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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Correct Answer: 1
Approach Solution - 1
To determine the molarity of the sodium iodide (NaI) solution, we'll follow these steps:
1. Write the Reaction Equation:
The reaction between sodium iodide (NaI) and silver nitrate (AgNO3) produces silver iodide (AgI) and sodium nitrate (NaNO3):
NaI + AgNO3 → AgI + NaNO3
2. Calculate Moles of Silver Iodide:
The molar mass of AgI = Ag + I = 108 + 127 = 235 g/mol. Given silver iodide mass is 4.74 g:
Moles of AgI = 4.74 g / 235 g/mol = 0.02017 mol
3. Relate Moles to Sodium Iodide:
From the balanced equation, moles of NaI = moles of AgI = 0.02017 mol
4. Calculate Molarity of NaI:
Volume of NaI solution = 20 mL = 0.020 L.
Molarity (M) = Moles of solute / Volume of solution in liters = 0.02017 mol / 0.020 L = 1.0085 M
Conclusion:
The nearest integer molarity of the sodium iodide solution is 1 M, which falls within the range of 1 to 1. Thus, the molarity is confirmed as 1 M.
Approach Solution -2
Step 2: Calculate moles of AgI precipitated Molar mass of AgI = \( 108 + 127 = 235 \) g mol\(^{-1}\) \[ \text{Moles of AgI} = \frac{4.74 \text{ g}}{235 \text{ g mol}^{-1}} = 0.0202 \text{ mol} \]
Step 3: Determine moles of NaI From stoichiometry, 1 mole NaI produces 1 mole AgI \[ \text{Moles of NaI} = 0.0202 \text{ mol} \]
Step 4: Calculate molarity of NaI solution \[ \text{Molarity} = \frac{\text{Moles}}{\text{Volume in L}} = \frac{0.0202 \text{ mol}}{0.020 \text{ L}} = 1.01 \text{ M} \] Rounding to nearest integer: \[ \text{Molarity} \approx 1 \text{ M} \]
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