Identify the correct statements: A. Hydrated salts can be used as primary standard. B. Primary standard should not undergo any reaction with air. C. Reactions of primary standard with another substance should be instantaneous and stoichiometric. D. Primary standard should not be soluble in water. E. Primary standard should have low relative molar mass. Choose the correct answer from the options given below:

Related Formula

Equivalents of Acid = Equivalents of Base

Core Logic

Write the balanced chemical equation to find the molar ratio, or use the concept of equivalents directly. Barium hydroxide is a diacidic base (n$n$-factor = 2) and hydrochloric acid is a monobasic acid (n$n$-factor = 1).

Ba(OH)2(aq) + 2HCl(aq) arrow BaCl2(aq) + 2H₂O(l)$Ba(OH)_{2(aq)} + 2HCl_{(aq)} \rightarrow BaCl_{2(aq)} + 2H_2O_{(l)}$

Step 1: Calculating Equivalents

Millimoles of Ba(OH)₂ = M × V = 0.1 M × 25.0 mL = 2.5 mmoles$Ba(OH)_2 = M \times V = 0.1 \text{ M} \times 25.0 \text{ mL} = 2.5 \text{ mmoles}$. Since 1 mole of Ba(OH)₂$Ba(OH)_2$ neutralizes 2 moles of HCl$HCl$: Millimoles of HCl = 2 × 2.5 = 5.0 mmoles$HCl = 2 \times 2.5 = 5.0 \text{ mmoles}$.

Step 2: Calculating Mass

Weight of HCl = mmoles × molar mass (mg/mmol)$HCl = \text{mmoles} \times \text{molar mass (mg/mmol)}$ Weight = 5.0 × 36.5 = 182.5 mg$5.0 \times 36.5 = 182.5 \text{ mg}$.

Step 3: Matching Format

We need the answer in the form of x × 10⁻¹$x \times 10^{-1}$ mg. 182.5 mg = 1825 × 10⁻¹ mg$182.5 \text{ mg} = 1825 \times 10^{-1} \text{ mg}$. Thus, x = 1825$x = 1825$.

Pattern Recognition

Always multiply the Molarity by the n-factor to get Normality when performing titrations. 0.1 M Ba(OH)₂ = 0.2 N$0.1 \text{ M } Ba(OH)_2 = 0.2 \text{ N}$. Equivalents = 0.2 × 25 = 5 meq$0.2 \times 25 = 5 \text{ meq}$. Mass = 5 × 36.5 = 182.5 mg$5 \times 36.5 = 182.5 \text{ mg}$.

Chapter Mix

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Core Logic

Step 1: Calculate moles of HCl available. Density of solution (d$d$) = 1.13 g/mL$= 1.13\text{ g/mL}$. Volume (V$V$) = 300 mL$= 300\text{ mL}$. Mass of solution = 300 × 1.13 = 339 g$= 300 \times 1.13 = 339\text{ g}$. Mass of pure HCl = 339 × (38.55)/(100) = 130.68 g$= 339 \times \frac{38.55}{100} = 130.68\text{ g}$. Molar mass of HCl = 36.5 g/mol$= 36.5\text{ g/mol}$. Moles of HCl initially = (130.68)/(36.5) = 3.58 moles$= \frac{130.68}{36.5} = 3.58\text{ moles}$.

Step 2: Calculate moles of CaCO₃$CaCO_3$ available. Molar mass of CaCO₃ = 100 g/mol$CaCO_3 = 100\text{ g/mol}$. Moles of CaCO₃ = (90)/(100) = 0.90 mole$CaCO_3 = \frac{90}{100} = 0.90\text{ mole}$.

Step 1: Determine Limiting Reagent and Reaction

Reaction: CaCO₃ + 2HCl arrow CaCl₂ + H₂O + CO₂$CaCO_3 + 2HCl \rightarrow CaCl_2 + H_2O + CO_2$ From the stoichiometry, 1 mole of CaCO₃$CaCO_3$ requires 2 moles of HCl$HCl$. 0.90 mole$0.90\text{ mole}$ of CaCO₃$CaCO_3$ requires 0.90 × 2 = 1.80 moles$0.90 \times 2 = 1.80\text{ moles}$ of HCl$HCl$. Since 3.58 moles of HCl$HCl$ are available, CaCO₃$CaCO_3$ is the limiting reagent (LR) and HCl$HCl$ is in excess.

Moles of HCl$HCl$ remained unreacted = 3.58 - 1.80 = 1.78 moles$= 3.58 - 1.80 = 1.78\text{ moles}$. Mass of HCl$HCl$ remained = 1.78 × 36.5 = 64.97 g$= 1.78 \times 36.5 = 64.97\text{ g}$.

Pattern Recognition

Always convert volume, density, and mass percentage into pure mass. Identify Limiting Reagent before finding the leftover.

Chapter Mix

Class 11 Chemistry: Some Basic Concepts of Chemistry

Related Formula

Molarity conversion relation matrix:

Core Logic

Let's perform molar quantities verification row-by-row:

• Molar mass properties: CaCO₃ = 100~g/mol$\mathrm{CaCO_3} = 100\mathrm{~g/mol}$, CaCl₂ = 40 + (35.5 × 2) = 111~g/mol$\mathrm{CaCl_2} = 40 + (35.5 \times 2) = 111\mathrm{~g/mol}$.
• Initial chemical moles calculated:

• Determine the limiting reactant via stoichiometric ratios: HCl$\mathrm{HCl}$ acts as the Limiting Reagent (L.R.) because its proportional structural requirement is much smaller.
• Moles of product CaCl₂$\mathrm{CaCl_2}$ formed based on L.R. configuration:

Pattern Recognition

Always compare the available moles divided by the respective stoichiometric coefficients to quickly find the Limiting Reagent: 10/1 gg 0.19/2$10/1 \gg 0.19/2$. This trick saves execution seconds during complex numeric problems.

Chapter Mix

Class 11 Chemistry: Some Basic Concepts of Chemistry

Related Formula

Number of millimoles (n$n$) is given by:

Molarity (M$M$) of a diluted mixture:

Core Logic

Evaluate the first mixing step to determine the net acid-base state:

• Millimoles of NaOH = 10~mL × 2~M = 20~mmol$\mathrm{NaOH} = 10\mathrm{~mL} \times 2\mathrm{~M} = 20\mathrm{~mmol}$
• Millimoles of HCl = 20~mL × 1~M = 20~mmol$\mathrm{HCl} = 20\mathrm{~mL} \times 1\mathrm{~M} = 20\mathrm{~mmol}$

Since millimoles are equal, HCl$\mathrm{HCl}$ and NaOH$\mathrm{NaOH}$ completely neutralize each other, producing a neutral aqueous salt solution.

Step 1: Analyze transfer to volumetric flask

Taking 10~mL$10\mathrm{~mL}$ of this neutralized solution provides no excess H^+$\mathrm{H}^+$ or OH^-$\mathrm{OH}^-$ ions. It is added to a volumetric flask containing 2~mol$2\mathrm{~mol}$ of pure HCl$\mathrm{HCl}$.

Step 2: Calculate final molarity of HCl$\mathrm{HCl}$

The volume of the flask is made up to 100~mL = 0.1~L$100\mathrm{~mL} = 0.1\mathrm{~L}$:

Hence, the resulting solution is 20~M~HCl$20\mathrm{~M~HCl}$.

Pattern Recognition

Stoichiometric neutralizations are evaluated by setting up mole/millimole balance charts. Once stoichiometric equivalence (MVacid = MVbase$MV_{\text{acid}} = MV_{\text{base}}$) is reached, any sub-aliquot of that solution remains completely neutral.

Chapter Mix

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