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A white solid is a mixture of sodium ethandioate (Na2C2O4), ethandioic acid dihydrate (H2C2O4·2H2O) and an inert solid - AQA - A-Level Chemistry - Question 11 - 2017 - Paper 1

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Question 11

A-white-solid-is-a-mixture-of-sodium-ethandioate-(Na2C2O4),-ethandioic-acid-dihydrate-(H2C2O4·2H2O)-and-an-inert-solid-AQA-A-Level Chemistry-Question 11-2017-Paper 1.png

A white solid is a mixture of sodium ethandioate (Na2C2O4), ethandioic acid dihydrate (H2C2O4·2H2O) and an inert solid. A volumetric flask contained 1.90 g of this s... show full transcript

Worked Solution & Example Answer:A white solid is a mixture of sodium ethandioate (Na2C2O4), ethandioic acid dihydrate (H2C2O4·2H2O) and an inert solid - AQA - A-Level Chemistry - Question 11 - 2017 - Paper 1

Step 1

Calculate moles of MnO4- in the first titration

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Answer

Using the volume and molarity of potassium manganate(VII):

Moles of KMnO4 = Molarity × Volume = 0.200 mol dm⁻³ × (26.50 cm³ / 1000) = 0.00530 mol.

From the balanced equation:

1 mole of MnO4- reacts with 5 moles of C2O4^2-, hence:

Moles of C2O4^2- = 5 × moles of MnO4- = 5 × 0.00530 = 0.0265 mol.

Step 2

Calculate moles of H2C2O4 in the second titration

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Answer

Using the volume and molarity of NaOH:

Moles of NaOH = 0.100 mol dm⁻³ × (10.45 cm³ / 1000) = 0.001045 mol.

From the balanced equation:

1 mole of H2C2O4 reacts with 2 moles of NaOH, hence:

Moles of H2C2O4 = 0.001045 mol / 2 = 0.0005225 mol.

Step 3

Calculate total moles of sodium ethandioate

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Answer

Since sodium ethandioate is formed from ethandioic acid:

Total moles of sodium ethandioate = moles of C2O4^2- + moles of H2C2O4 = 0.0265 + 0.0005225 = 0.0270225 mol.

Step 4

Calculate mass of sodium ethandioate in the solid mixture

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Answer

Molar mass of sodium ethandioate (Na2C2O4) = (2 × 23) + (2 × 12) + (4 × 16) = 110 g mol⁻¹.

Mass = moles × molar mass = 0.0270225 mol × 110 g mol⁻¹ = 2.9725 g.

Step 5

Calculate percentage by mass of sodium ethandioate

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Answer

Percentage by mass = (mass of sodium ethandioate / total mass of the solid) × 100 = (2.9725 g / 1.90 g) × 100 = 156.5 %.

Since this value exceeds 100%, revise assumptions and check calculations.

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