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5g of mixture of NaOH and KOH were dissolved and made upto 250 ml .25 ml of solution were completely neutralised by 17ml of N/2 HCl solution . The % of KOH in mixture .?
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5g of mixture of NaOH and KOH were dissolved and made upto 250 ml .25 ...
Introduction
To find the percentage of KOH in the mixture of NaOH and KOH, we begin by analyzing the neutralization reaction with HCl.
Step 1: Calculate Moles of HCl
- The normality of HCl solution is N/2.
- Volume of HCl used = 17 mL = 0.017 L.
- Moles of HCl = Normality × Volume (in L) = (0.5) × (0.017) = 0.0085 moles.
Step 2: Determine Moles of Bases Neutralized
- Since NaOH and KOH both react with HCl in a 1:1 ratio, total moles of NaOH and KOH in 25 mL of the solution is also 0.0085 moles.
Step 3: Calculate Concentration in 250 mL
- The total volume of the solution is 250 mL, so concentration in 1 L = (0.0085 moles in 25 mL) × (1000/25) = 0.34 moles/L.
Step 4: Set Up the Equation
- Let x be the grams of KOH and (5 - x) be the grams of NaOH.
- Molar mass of KOH = 56.11 g/mol, NaOH = 40 g/mol.
- Moles of KOH = x / 56.11, Moles of NaOH = (5 - x) / 40.
Step 5: Formulate the Equation
- Total moles: (x / 56.11) + ((5 - x) / 40) = 0.0085.
Step 6: Solve for x
- By solving the equation, we find the value of x (grams of KOH).
Step 7: Calculate Percentage of KOH
- Percentage of KOH = (grams of KOH / total grams of mixture) × 100.
Following these steps, you will determine the percentage of KOH in the mixture accurately. This method ensures a clear understanding of the neutralization concept and stoichiometry involved in the calculations.
To find the percentage of KOH in the mixture of NaOH and KOH, we begin by analyzing the neutralization reaction with HCl.
Step 1: Calculate Moles of HCl
- The normality of HCl solution is N/2.
- Volume of HCl used = 17 mL = 0.017 L.
- Moles of HCl = Normality × Volume (in L) = (0.5) × (0.017) = 0.0085 moles.
Step 2: Determine Moles of Bases Neutralized
- Since NaOH and KOH both react with HCl in a 1:1 ratio, total moles of NaOH and KOH in 25 mL of the solution is also 0.0085 moles.
Step 3: Calculate Concentration in 250 mL
- The total volume of the solution is 250 mL, so concentration in 1 L = (0.0085 moles in 25 mL) × (1000/25) = 0.34 moles/L.
Step 4: Set Up the Equation
- Let x be the grams of KOH and (5 - x) be the grams of NaOH.
- Molar mass of KOH = 56.11 g/mol, NaOH = 40 g/mol.
- Moles of KOH = x / 56.11, Moles of NaOH = (5 - x) / 40.
Step 5: Formulate the Equation
- Total moles: (x / 56.11) + ((5 - x) / 40) = 0.0085.
Step 6: Solve for x
- By solving the equation, we find the value of x (grams of KOH).
Step 7: Calculate Percentage of KOH
- Percentage of KOH = (grams of KOH / total grams of mixture) × 100.
Following these steps, you will determine the percentage of KOH in the mixture accurately. This method ensures a clear understanding of the neutralization concept and stoichiometry involved in the calculations.
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5g of mixture of NaOH and KOH were dissolved and made upto 250 ml .25 ml of solution were completely neutralised by 17ml of N/2 HCl solution . The % of KOH in mixture .?
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5g of mixture of NaOH and KOH were dissolved and made upto 250 ml .25 ml of solution were completely neutralised by 17ml of N/2 HCl solution . The % of KOH in mixture .? for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about 5g of mixture of NaOH and KOH were dissolved and made upto 250 ml .25 ml of solution were completely neutralised by 17ml of N/2 HCl solution . The % of KOH in mixture .? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 5g of mixture of NaOH and KOH were dissolved and made upto 250 ml .25 ml of solution were completely neutralised by 17ml of N/2 HCl solution . The % of KOH in mixture .?.
5g of mixture of NaOH and KOH were dissolved and made upto 250 ml .25 ml of solution were completely neutralised by 17ml of N/2 HCl solution . The % of KOH in mixture .? for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about 5g of mixture of NaOH and KOH were dissolved and made upto 250 ml .25 ml of solution were completely neutralised by 17ml of N/2 HCl solution . The % of KOH in mixture .? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 5g of mixture of NaOH and KOH were dissolved and made upto 250 ml .25 ml of solution were completely neutralised by 17ml of N/2 HCl solution . The % of KOH in mixture .?.
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