Concentrated nitric acid is labelled as 75%$75\%$ by mass. The volume in mL of the solution which contains 30 g$30\mathrm{\ g}$ of nitric acid is Given: Density of nitric acid solution is 1.25 g/mL$1.25\mathrm{\ g/mL}$

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.

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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.

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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