Solution A: Sulfuric Acid (H2SO4)

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 ($H_2SO_4$)

The instruction states that the solution should contain 5.60 $cm^3$ of concentrated $H_2SO_4$ (R.D. 1.84 and purity 98%) per $dm^3$ of solution. Each candidate receives 150 $cm^3$ of this solution.

Given: • Volume of concentrated $H_2SO_4$ per $dm^3$ of solution = 5.60 $cm^3$ • Relative Density (R.D.) of concentrated $H_2SO_4$ = 1.84 • Purity of concentrated $H_2SO_4$ = 98% • Molar mass of $H_2SO_4$ = 98.08 g/mol

Step 1: Calculate the density of concentrated $H_2SO_4$. Assuming the density of water is 1.00 g/$cm^3$, the density of concentrated $H_2SO_4$ is: $Density = R.D. \times Density of water = 1.84 \times 1.00 g/cm^3 = 1.84 g/cm^3$

Step 2: Calculate the mass of concentrated $H_2SO_4$ in 5.60 $cm^3$. $Mass of conc. H_2SO_4 = Volume \times Density = 5.60 cm^3 \times 1.84 g/cm^3 = 10.304 g$

Step 3: Calculate the mass of pure $H_2SO_4$ in this amount, considering its purity. $Mass of pure H_2SO_4 = Mass of conc. H_2SO_4 \times \frac{Purity}{100} = 10.304 g \times \frac{98}{100} = 10.09792 g$

Step 4: Calculate the moles of pure $H_