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Some Basic Concepts of Chemistry question

2025 · 8 Apr · Shift 2 · Q24
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Some Basic Concepts of Chemistry question

2025 · 8 Apr · Shift 2 · Q24

JEE MainChemistrySome Basic Concepts of ChemistryNumerical+4 / −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 ‾\underline{\hspace{2cm}}​ M. (Nearest Integer value) (Given : Na = 23, I = 127, Ag = 108, N = 14, O = 16 g mol-1)
Numerical answer
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Correct answer: 1

  1. Write the reaction

    Sodium iodide reacts with silver nitrate as: NaI+AgNO3→AgI↓+NaNO3\text{NaI} + \text{AgNO}_3 \rightarrow \text{AgI} \downarrow + \text{NaNO}_3NaI+AgNO3​→AgI↓+NaNO3​

    From the equation, the mole ratio is: NaI:AgI=1:1\text{NaI} : \text{AgI} = 1:1NaI:AgI=1:1

  2. Find molar mass of silver iodide

    M(AgI)=M(Ag)+M(I)=108+127=235 g mol−1M(\text{AgI}) = M(\text{Ag}) + M(\text{I}) = 108 + 127 = 235\ \text{g mol}^{-1}M(AgI)=M(Ag)+M(I)=108+127=235 g mol−1

  3. Calculate moles of AgI formed

    Given mass of AgI = 4.74 g4.74\ \text{g}4.74 g

    n(AgI)=4.74235=0.02017 moln(\text{AgI}) = \frac{4.74}{235} = 0.02017\ \text{mol}n(AgI)=2354.74​=0.02017 mol

  4. Use stoichiometry to find moles of NaI

    Since the ratio is 1:11:11:1: n(NaI)=n(AgI)=0.02017 moln(\text{NaI}) = n(\text{AgI}) = 0.02017\ \text{mol}n(NaI)=n(AgI)=0.02017 mol

  5. Convert volume into litres

    Given volume of NaI solution = 20 mL20\ \text{mL}20 mL 20 mL=0.020 L20\ \text{mL} = 0.020\ \text{L}20 mL=0.020 L

  6. Calculate molarity

    M=moles of solutevolume in litresM = \frac{\text{moles of solute}}{\text{volume in litres}}M=volume in litresmoles of solute​

    M=0.020170.020=1.0085 MM = \frac{0.02017}{0.020} = 1.0085\ \text{M}M=0.0200.02017​=1.0085 M

  7. Nearest integer value

    1.0085≈11.0085 \approx 11.0085≈1

Final Answer: 111

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