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A volume of 10ml H2S04 solution is diluted to 100ml. Twenty five ml of this diluted solution is mixed with 50ml of 0.5 N NaOH solution. The resulting solution requires 0.265 g Na2CO3 for complete neutralization. The normality of original H2SO4 solution is. a)12N b)11N c) 3N d) 0.275 N?
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A volume of 10ml H2S04 solution is diluted to 100ml. Twenty five ml of...
Solution:
Given,
Volume of H2SO4 solution = 10 ml
Volume of diluted H2SO4 solution = 100 ml
Volume of diluted H2SO4 solution used = 25 ml
Volume of NaOH solution = 50 ml
Amount of Na2CO3 required for complete neutralization = 0.265 g

Let the normality of original H2SO4 solution be x N.

Calculations:

1. Calculation of Normality of NaOH Solution:

Molarity of NaOH solution = 0.5 N
Molarity of NaOH solution = Normality of NaOH solution × Equivalent weight of NaOH
Equivalent weight of NaOH = Molecular weight of NaOH / Number of H+ ions replaced
Molecular weight of NaOH = 40 g/mol
Number of H+ ions replaced by NaOH = 1
Equivalent weight of NaOH = 40 g/mol
0.5 N = Normality of NaOH × 40 g/mol
Normality of NaOH = 0.5 / 40
Normality of NaOH = 0.0125 N

2. Calculation of Number of moles of NaOH used:

Number of moles of NaOH = Molarity of NaOH × Volume of NaOH solution (in L)
Number of moles of NaOH = 0.0125 × (50/1000)
Number of moles of NaOH = 0.000625 mol

3. Calculation of Number of moles of H2SO4 in 25 ml of diluted H2SO4 solution:

Molarity of diluted H2SO4 solution = x N
Molarity of diluted H2SO4 solution = Number of moles of H2SO4 / Volume of diluted H2SO4 solution (in L)
Number of moles of H2SO4 = Molarity of diluted H2SO4 solution × Volume of diluted H2SO4 solution (in L)
Number of moles of H2SO4 = x × (25/1000)
Number of moles of H2SO4 = 0.025x mol

4. Calculation of Number of moles of H2SO4 in 100 ml of diluted H2SO4 solution:

Number of moles of H2SO4 in 100 ml of diluted H2SO4 solution = Number of moles of H2SO4 in 25 ml of diluted H2SO4 solution × 4
Number of moles of H2SO4 in 100 ml of diluted H2SO4 solution = 0.1x mol

5. Calculation of Number of moles of Na2CO3 used:

Number of moles of Na2CO3 = Mass of Na2CO3 / Molecular weight of Na2CO3
Molecular weight of Na2CO3 = 106 g/mol
Number of moles of Na2CO3 = 0.265 / 106
Number of moles of Na2CO3 = 0.0025 mol

6. Calculation of Number of moles of H2SO4 required for complete neutralization:

The balanced chemical equation for the reaction between Na2CO3 and H2SO4 is:

Na2CO3 + H2SO4 → Na2SO4 + H2O + CO2

From the equation, it
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A volume of 10ml H2S04 solution is diluted to 100ml. Twenty five ml of...
11N
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A volume of 10ml H2S04 solution is diluted to 100ml. Twenty five ml of this diluted solution is mixed with 50ml of 0.5 N NaOH solution. The resulting solution requires 0.265 g Na2CO3 for complete neutralization. The normality of original H2SO4 solution is. a)12N b)11N c) 3N d) 0.275 N?
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A volume of 10ml H2S04 solution is diluted to 100ml. Twenty five ml of this diluted solution is mixed with 50ml of 0.5 N NaOH solution. The resulting solution requires 0.265 g Na2CO3 for complete neutralization. The normality of original H2SO4 solution is. a)12N b)11N c) 3N d) 0.275 N? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about A volume of 10ml H2S04 solution is diluted to 100ml. Twenty five ml of this diluted solution is mixed with 50ml of 0.5 N NaOH solution. The resulting solution requires 0.265 g Na2CO3 for complete neutralization. The normality of original H2SO4 solution is. a)12N b)11N c) 3N d) 0.275 N? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A volume of 10ml H2S04 solution is diluted to 100ml. Twenty five ml of this diluted solution is mixed with 50ml of 0.5 N NaOH solution. The resulting solution requires 0.265 g Na2CO3 for complete neutralization. The normality of original H2SO4 solution is. a)12N b)11N c) 3N d) 0.275 N?.
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