A 20 mL$20 \text{ mL}$ sample of a sodium iodide solution yields 4.74 g$4.74 \text{ g}$ of silver iodide precipitate when treated with an excess of silver nitrate solution. The molarity of the initial sodium iodide solution is _________ M (as the nearest integer value). Given molar masses: Na = 23, I = 127, Ag = 108, N = 14, O = 16 g mol⁻¹$\text{Na} = 23, \, \text{I} = 127, \, \text{Ag} = 108, \, \text{N} = 14, \, \text{O} = 16 \text{ g mol}^{-1}$.

Related Formula

The number of atoms in a given mass of an element is calculated using:

Core Logic

Since the mass (10⁻⁹ g$10^{-9}\text{ g}$) is identical for all samples, the number of atoms is inversely proportional to the molar mass of the element:

Step 1: Molar Mass Values Comparison

Let us check the molar masses of the listed elements:

• Molar Mass of Po ≈ 209 g/mol$\text{Molar Mass of Po} \approx 209\text{ g/mol}$
• Molar Mass of Pr ≈ 141 g/mol$\text{Molar Mass of Pr} \approx 141\text{ g/mol}$
• Molar Mass of Pb ≈ 207 g/mol$\text{Molar Mass of Pb} \approx 207\text{ g/mol}$
• Molar Mass of Pt ≈ 195 g/mol$\text{Molar Mass of Pt} \approx 195\text{ g/mol}$

Step 2: Conclusion

Praseodymium (Pr) has the least molar mass (141 g/mol$141\text{ g/mol}$), meaning it will yield the maximum total number of atoms for the specified mass sample.

Pattern Recognition

Shortcut: Equal mass given arrow$\rightarrow$ Lighter atoms mean more atoms per gram. Find the element with the lowest atomic mass value from the choices.

Evaluation Rubric / Model Answer

Option (B)

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Class 11 Chemistry: Some Basic Concepts of Chemistry

Related Formula

Core Logic

• Find the pure mass of calcium carbonate (CaCO₃$CaCO_3$) present:

• Convert this mass into moles (Molar mass of CaCO₃ = 40+12+48 = 100 ~g· mol⁻¹$CaCO_3 = 40+12+48 = 100 \mathrm{~g\cdot mol^{-1}}$):

• From the stoichiometry of the reaction, 1 mole of CaCO₃$CaCO_3$ yields 1 mole of CaO$CaO$:

Pattern Recognition

Always multiply by the purity fraction first before entering regular stoichiometric conversion chains. Since the formula weight of limestone is exactly 100$100$, tracking percentages directly mirrors mole factors seamlessly.

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Class 11 Chemistry: Some Basic Concepts of Chemistry

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Core Logic

Let the required mass of pure iron be w~g$w\mathrm{~g}$. The total mass of the wheat mixture is 150~kg = 150 × 10³~g$150\mathrm{~kg} = 150 \times 10^3\mathrm{~g}$. Applying the parts-per-million concentration condition:

Now, determine the molar mass of the complete green vitriol salt crystal template, FeSO₄ · 7H₂O$\mathrm{FeSO_4 \cdot 7H_2O}$:

Set up a stoichiometric mass balance proportion to find the total salt mass w₁$w_1$:

Rounding off to the nearest integer value gives 9.

Pattern Recognition

Always convert concentration metrics back to absolute molar mass equivalence values before distributing across full hydrated molecular templates.

Chapter Mix

Class 11 Chemistry: Some Basic Concepts of Chemistry

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Core Logic

First, calculate initial molar quantities for both reactants:

• Molar mass of C₄H₁₀ = 4(12) + 10(1) = 58 g/mol$\text{C}_4\text{H}_{10} = 4(12) + 10(1) = 58\text{ g/mol}$
• Initial Moles of butane = 174.0 × 10³ g58 g/mol = 3000 mol = 3 × 10³ mol$= \frac{174.0 \times 10^3\text{ g}}{58\text{ g/mol}} = 3000\text{ mol} = 3 \times 10^3\text{ mol} $
• Molar mass of O₂ = 32 g/mol$\text{O}_2 = 32\text{ g/mol}$
• Initial Moles of oxygen = 320.0 × 10³ g32 g/mol = 10000 mol = 10 × 10³ mol$= \frac{320.0 \times 10^3\text{ g}}{32\text{ g/mol}} = 10000\text{ mol} = 10 \times 10^3\text{ mol} $

Step 1: Identify Limiting Reagent

Let's test the stoichiometric requirements via calculation ratios:

• Ratio for C₄H₁₀ = (3000)/(1) = 3000$\text{C}_4\text{H}_{10} = \frac{3000}{1} = 3000$
• Ratio for O₂ = (10000)/(13/2) = 1538.46$\text{O}_2 = \frac{10000}{13/2} = 1538.46$

Since the ratio for O₂$\text{O}_2$ is lower, oxygen behaves as the limiting reagent and commands the output steps.

Step 2: Compute Water Yield

Using stoichiometric proportions defined by the balanced reaction field:

Convert moles to mass (MH2O = 18 g/mol$M{\text{H}2\text{O}} = 18\text{ g/mol}$):

Since density = 1 g/mL = 1 kg/L$= 1\text{ g/mL} = 1\text{ kg/L}$, the net volume is exactly:

Pattern Recognition

Limiting reagent shortcut: Always check ratios (moles / stoichiometric coefficient) right away. Do not spend time calculating theoretical products based on butane before establishing whether oxygen runs out first.

Chapter Mix

Class 11 Chemistry: Some Basic Concepts of Chemistry

Core Logic

Let's find the empirical formula by calculating the relative molar ratios of each element:

• Carbon (C):

• Hydrogen (H):

• Oxygen (O):

Next, divide each value by the smallest number of moles (2.287$2.287$) to get the simplest whole-number ratio:

• C: (4.516)/(2.287) ≈ 1.97 ≈ 2$\frac{4.516}{2.287} \approx 1.97 \approx 2$
• H: (9.200)/(2.287) ≈ 4.02 ≈ 4$\frac{9.200}{2.287} \approx 4.02 \approx 4$
• O: (2.287)/(2.287) = 1$\frac{2.287}{2.287} = 1$

Thus, the Empirical Formula is C₂H₄O$\mathrm{C}_{2}\mathrm{H}_{4}\mathrm{O}$.

Step 2: Molecular Formula Determination

• Calculate the empirical formula mass:

• Find the multiplier (n$n$):

• Determine the molecular formula:

This matches Option (3).

Pattern Recognition

Always determine the empirical formula ratio first. Then, compare the empirical mass with the given molar mass to find the whole-number multiplier (n$n$). Multiplication gives the definitive molecular formula.

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Class 11 Chemistry: Some Basic Concepts of Chemistry