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Solution A: Sulfuric Acid (H2SO4)

Chemistry

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Here are the calculations for the concentrations of solutions A and B, and the description for mixture C, based on the provided instructions.

Solution A: Sulfuric Acid (H2SO4H_2SO_4)

The instruction states that the solution should contain 5.60 cm3cm^3 of concentrated H2SO4H_2SO_4 (R.D. 1.84 and purity 98%) per dm3dm^3 of solution. Each candidate receives 150 cm3cm^3 of this solution.

Given:

  • Volume of concentrated H2SO4H_2SO_4 per dm3dm^3 of solution = 5.60 cm3cm^3
  • Relative Density (R.D.) of concentrated H2SO4H_2SO_4 = 1.84
  • Purity of concentrated H2SO4H_2SO_4 = 98%
  • Molar mass of H2SO4H_2SO_4 = 98.08 g/mol

  1. Calculate the density of concentrated H2SO4H_2SO_4.

    Assuming the density of water is 1.00 g/cm3cm^3, the density of concentrated H2SO4H_2SO_4 is:

    Density=R.D.×Density of water=1.84×1.00 g/cm3=1.84 g/cm3\text{Density} = \text{R.D.} \times \text{Density of water} = 1.84 \times 1.00 \text{ g/cm}^3 = 1.84 \text{ g/cm}^3

  2. Calculate the mass of concentrated H2SO4H_2SO_4 in 5.60 cm3cm^3.

    Mass of conc. H2SO4=Volume×Density=5.60 cm3×1.84 g/cm3=10.304 g\text{Mass of conc. } H_2SO_4 = \text{Volume} \times \text{Density} = 5.60 \text{ cm}^3 \times 1.84 \text{ g/cm}^3 = 10.304 \text{ g}

  3. Calculate the mass of pure H2SO4H_2SO_4 in this amount, considering its purity.

    Mass of pure H2SO4=Mass of conc. H2SO4×Purity100=10.304 g×98100=10.09792 g\text{Mass of pure } H_2SO_4 = \text{Mass of conc. } H_2SO_4 \times \frac{\text{Purity}}{100} = 10.304 \text{ g} \times \frac{98}{100} = 10.09792 \text{ g}

  4. Calculate the moles of pure $H

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