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Question

An element X forms a chloride with the formula XCl$_3$ and another chloride with the formula XCl$_5$. 0.1 moles of XCl$_3$ requires 60 mL of 1 N NaOH for complete hydrolysis, while 0.1 moles of XCl$_5$ requires 100 mL of 1 N NaOH. What is the equivalent weight of the element X in XCl$_3$ and XCl$_5$ respectively?

Mr. S M HUSSAIN · Accepted Answer

## CORE INFORMATION\n✓ **Answer**: B\n## CONCEPT FRAMEWORK\n\t**Primary Concept**: Equivalent weight and Hydrolysis of Chlorides\n\t**Related Topics**: Stoichiometry, Normality, Acidity of Chlorides\n\t**Prerequisites**: Understanding of molarity, normality, and equivalent weight.\n## SOLUTION PATHWAY\n**Given Information**:\n\t- XCl$_3$ and XCl$_5$ are hydrolyzed by NaOH.\n\t- 0.1 moles of XCl$_3$ reacts with 60 mL of 1 N NaOH.\n\t- 0.1 moles of XCl$_5$ reacts with 100 mL of 1 N NaOH.\n**Method & Steps**:\n\t1. **Calculate the milliequivalents of NaOH reacted in each case:**\n\t\tReasoning: Milliequivalents of NaOH = Normality * Volume (mL)\n\t\tFormula/Rule: Milliequivalents = N * V\n\t2. **Determine the milliequivalents of XCl$_3$ and XCl$_5$:**\n\t\tWorking:  Milliequivalents of XCl$_3$ = Milliequivalents of NaOH = 1 N * 60 mL = 60 milliequivalents. Similarly, milliequivalents of XCl$_5$ = 1 N * 100 mL = 100 milliequivalents.\n\t\tKey Point: In a complete reaction, milliequivalents of reactants are equal to milliequivalents of products.\n\t3. **Calculate the equivalent weight of X in each compound:**\n\t\tReasoning: Equivalent weight = (Moles * Molar mass of the compound) / Number of equivalents\n\t\tWorking: Since we have 0.1 moles of each compound, we can find the number of equivalents by dividing the milliequivalents by 1000.  So, for XCl$_3$:  Equivalent weight of X = (0.1 * Molar mass of XCl$_3$) / (60/1000) = (Molar mass of XCl$_3$)/6.  Since the equivalent weight of Cl is its atomic weight, the equivalent weight of X in XCl$_3$ is (Molar mass of X)/6. For XCl$_5$: Equivalent weight of X = (0.1 * Molar mass of XCl$_5$) / (100/1000) = (Molar mass of XCl$_5$)/10.  The equivalent weight of X in XCl$_5$ is (Molar mass of X)/10.\n\t\tResult: (Molar mass of X)/6 and (Molar mass of X)/10\n\t\tVerification: The equivalent weight is inversely proportional to the amount of NaOH required for hydrolysis.  Since XCl$_5$ requires more NaOH, its equivalent weight is lower, which is consistent with the result.\n## OPTION ANALYSIS\n[A] :x: Incorrect. This assumes a 1:1 reaction with NaOH, which is not the case during hydrolysis.\n[B] ✓: Correct. As explained in the solution pathway.\n[C] :x: Incorrect. This doesn't account for the different number of chlorine atoms and their contribution to the hydrolysis reaction.\n[D] :x: Incorrect. This doesn't accurately reflect the stoichiometry of the hydrolysis reactions.\n## LEARNING AIDS\n\t**Common Mistakes**:\n\t\t1. Confusing molarity with normality: Remember that normality accounts for the reactive capacity of a substance.\n\t\t2. Incorrectly balancing the hydrolysis reaction: Properly balance the reaction to determine the correct stoichiometry.\n\t**Quick Tips**:\n\t\t1. Remember the relationship between milliequivalents: Milliequivalents of acid = Milliequivalents of base in a neutralization reaction.\n\t\t2. Equivalent weight of an element in a compound is related to its oxidation state change during the reaction.

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Mr. S M HUSSAIN · Answer

(Molar mass of X)/3 and (Molar mass of X)/5

Mr. S M HUSSAIN · Answer

The correct Option is "(Molar mass of X)/6 and (Molar mass of X)/10"

Mr. S M HUSSAIN · Answer

(Molar mass of X)/2 and (Molar mass of X)/3

Mr. S M HUSSAIN · Answer

(Molar mass of X)/4 and (Molar mass of X)/6

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