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1.2 g of a salt with its empirical formula KxHy(C2O4)z was dissolved in 50 mL of water and its 10 mL portion required 11 mL of a 0.1 M HCl solution to reach the equivalence point. In a separate titration, 15 mL of the stock solution required 20 mL 0.2475 M KOH to reach the equivalence point. Identify the correct option.
  • a)
    y = 3x
  • b)
    y = 2x
  • c)
    y = 4x
  • d)
    y = 5x
Correct answer is option 'A'. Can you explain this answer?
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1.2 g of a salt with its empirical formula KxHy(C2O4)zwas dissolved in...
Given information:
- 1.2 g of a salt with empirical formula KxHy(C2O4)z dissolved in 50 mL of water
- 10 mL of the solution required 11 mL of 0.1 M HCl to reach equivalence point
- 15 mL of the solution required 20 mL of 0.2475 M KOH to reach equivalence point

To find: Empirical formula of the salt

Solution:
Step 1: Find the molarity of the salt solution
Molarity of HCl = 0.1 M
Volume of HCl used = 11 mL = 0.011 L
Moles of HCl used = Molarity x Volume = 0.1 x 0.011 = 0.0011 moles
Since HCl and the salt react in a 1:1 ratio, the moles of the salt present in 10 mL of the solution = 0.0011 moles
Molarity of the salt solution = Moles of the salt / Volume of the solution in L = 0.0011 / 0.01 = 0.11 M

Step 2: Find the number of moles of KOH used in the titration
Molarity of KOH = 0.2475 M
Volume of KOH used = 20 mL = 0.02 L
Moles of KOH used = Molarity x Volume = 0.2475 x 0.02 = 0.00495 moles

Step 3: Find the number of moles of the salt present in 15 mL of the solution
Since KOH and the salt react in a 1:1 ratio, the moles of the salt present in 15 mL of the solution = 0.00495 moles

Step 4: Find the empirical formula of the salt
The ratio of K to C2O4 in the salt can be found by comparing the moles of K and C2O4 in the salt.
Molar mass of K = 39 g/mol
Molar mass of C2O4 = 88 g/mol
Moles of K in 0.00495 moles of the salt = 0.00495 x x = 0.00495x
Moles of C2O4 in 0.00495 moles of the salt = 0.00495 x z
The ratio of K:C2O4 in the salt = (0.00495x/39) : (0.00495z/88) = (0.001x/39) : (0.001z/88) = (x/39) : (z/88)

Similarly, the ratio of H to C2O4 in the salt can be found by comparing the moles of H and C2O4 in the salt.
Molar mass of H = 1 g/mol
Moles of H in 0.00495 moles of the salt = 0.00495 y
The ratio of H:C2O4 in the salt = (0.00495y/1) : (0.00495z/88) = (y/1) : (z/88)

Since the empirical formula should have the smallest possible integers for x, y, and z, we can assume x = 1
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Community Answer
1.2 g of a salt with its empirical formula KxHy(C2O4)zwas dissolved in...




∴ y = 3x.
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Volumetric analysis is based on the principle of equivalence a that minloces lead in the ratio of their equivalents. At the equivalence point of the reaction, involving the reactahts A and B : Number of gram equivalents of A = Number of gram equivalents of B..If VA ml of solution A having normality NA react just completely with VB ml of solution B having normalityNB thenEquation(1), called normality equation is very useful in numerical of volumetric analysis. Equivalent masses of different substances :Basically the equivalent mass of a substance is defined as the pads by seam ctituhich combine with or displace 1.0078 parts (1part) by mass of hydrogen, 8 parts by mansetaisiosaid 35.5 parts by mass of chlorine.Mass of a substance expressed in gram equal to its equivalent man is gam equivalent mass. Equivalent mass of a substance is not constant but depends upon the maw i malice the substance participates.Equivalent mass of an acid in acid-base reaction is its mass in grannehidicodoinisooleaf replaceable Fi ions (= 1.0078 g1g).0n the other hand, equivalent mass of a bagels istolonellfacontairis 1 mole of replaceable 0H ions. 1 g equivalent mass each of an acid and base an ionclinegins saint and 1 mole of water= 18 g). Equivalent mass of an oxidising agent is its mass which gains 1 mole at ellecismitcan be obtained by dividing the molecular mass or formula mass by the total decrease in aididditaiiilberef one or more elements per molecule.0n the other hand, equivalent mass of a reducing agent is the mass of the substance which loses 1 mole of electorns. It can be calculated by dividing the molecular or formula mass of the subtance by the total increase in oxidation number of one or more elements per molecular or formula mass.Q.Which of the following solutions, when mixed with 100 ml of 0.05 M NaOH, will give a neutral solution?

Volumetric analysis is based on the principle of equivalence a that minloces lead in the ratio of their equivalents. At the equivalence point of the reaction, involving the reactahts A and B : Number of gram equivalents of A = Number of gram equivalents of B..If VA ml of solution A having normality NA react just completely with VB ml of solution B having normalityNB thenEquation(1), called normality equation is very useful in numerical of volumetric analysis. Equivalent masses of different substances :Basically the equivalent mass of a substance is defined as the pads by seam ctituhich combine with or displace 1.0078 parts (1part) by mass of hydrogen, 8 parts by mansetaisiosaid 35.5 parts by mass of chlorine.Mass of a substance expressed in gram equal to its equivalent man is gam equivalent mass. Equivalent mass of a substance is not constant but depends upon the maw i malice the substance participates.Equivalent mass of an acid in acid-base reaction is its mass in grannehidicodoinisooleaf replaceable Fi ions (= 1.0078 g1g).0n the other hand, equivalent mass of a bagels istolonellfacontairis 1 mole of replaceable 0H ions. 1 g equivalent mass each of an acid and base an ionclinegins saint and 1 mole of water= 18 g). Equivalent mass of an oxidising agent is its mass which gains 1 mole at ellecismitcan be obtained by dividing the molecular mass or formula mass by the total decrease in aididditaiiilberef one or more elements per molecule.0n the other hand, equivalent mass of a reducing agent is the mass of the substance which loses 1 mole of electorns. It can be calculated by dividing the molecular or formula mass of the subtance by the total increase in oxidation number of one or more elements per molecular or formula mass.Q.How many gram of phosphoric acid (H3PO4) would be required to neutralize 58 g of magnesium hydroxide?

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1.2 g of a salt with its empirical formula KxHy(C2O4)zwas dissolved in 50 mL of water and its 10 mL portion required 11 mL of a 0.1 M HCl solution to reach the equivalence point. In a separate titration, 15 mL of the stock solution required 20 mL 0.2475 M KOH to reach the equivalence point. Identify the correct option.a)y = 3xb)y = 2xc)y = 4xd)y = 5xCorrect answer is option 'A'. Can you explain this answer?
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1.2 g of a salt with its empirical formula KxHy(C2O4)zwas dissolved in 50 mL of water and its 10 mL portion required 11 mL of a 0.1 M HCl solution to reach the equivalence point. In a separate titration, 15 mL of the stock solution required 20 mL 0.2475 M KOH to reach the equivalence point. Identify the correct option.a)y = 3xb)y = 2xc)y = 4xd)y = 5xCorrect answer is option 'A'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about 1.2 g of a salt with its empirical formula KxHy(C2O4)zwas dissolved in 50 mL of water and its 10 mL portion required 11 mL of a 0.1 M HCl solution to reach the equivalence point. In a separate titration, 15 mL of the stock solution required 20 mL 0.2475 M KOH to reach the equivalence point. Identify the correct option.a)y = 3xb)y = 2xc)y = 4xd)y = 5xCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 1.2 g of a salt with its empirical formula KxHy(C2O4)zwas dissolved in 50 mL of water and its 10 mL portion required 11 mL of a 0.1 M HCl solution to reach the equivalence point. In a separate titration, 15 mL of the stock solution required 20 mL 0.2475 M KOH to reach the equivalence point. Identify the correct option.a)y = 3xb)y = 2xc)y = 4xd)y = 5xCorrect answer is option 'A'. Can you explain this answer?.
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