To solve this problem, we need to follow several steps:
1. Write the balanced equation for the reaction between the divalent metal M and sulfuric(VI) acid (H2SO4).
M + H2SO4 -> MSO4 + H2
2. Convert the given mass of M to moles. Since M is a divalent metal, the molar mass of M will be twice the relative atomic mass.
Molar mass of M = 0.24 g / 0.5 mol/dm³ = 0.48 g/mol
3. Determine the moles of M by dividing the mass by the molar mass:
Moles of M = 0.24 g / 0.48 g/mol = 0.5 mol
4. Determine the moles of H2SO4 used in the reaction by using the stoichiometry from the balanced equation. From the balanced equation, we can see that the ratio between M and H2SO4 is 1:1.
Moles of H2SO4 = 0.5 mol
5. Determine the volume of 1.0M NaOH solution required for complete neutralization.
Moles of NaOH = moles of H2SO4 = 0.5 mol
Volume of NaOH = 5.0 cm³
6. From the concentration of the NaOH solution, calculate the moles of NaOH in 5.0 cm³.
Concentration of NaOH = 1.0 mol/dm³
Moles of NaOH = concentration × volume / 1000 = 1.0 mol/dm³ × 5.0 cm³ / 1000 cm³/dm³ = 0.005 mol
7. Since the ratio between NaOH and H2SO4 is 2:1 from the balanced equation, the moles of H2SO4 will be half the moles of NaOH.
Moles of H2SO4 = 0.005 mol / 2 = 0.0025 mol
8. Now, we can determine the initial concentration of H2SO4 using the volume and moles of H2SO4.
Concentration of H2SO4 = moles of H2SO4 / volume = 0.0025 mol / 500 cm³ = 0.005 mol/dm³ = 0.005 M
9. Finally, since 1 mole of M reacts with 1 mole of H2SO4, the molar ratio between M and H2SO4 is 1:1.
Therefore, the relative atomic mass of M can be calculated as follows:
Relative atomic mass of M = Molar mass of M / Moles of M = 0.48 g/mol / 0.5 mol = 0.96 g/mol
Therefore, the relative atomic mass of M is 0.96 g/mol.
0.24g of a divalent metal M dissolve in 500cm³ of 0.25M sulphuric(VI)acid.The resulting solution required solution required 5.0cm³ of 1.0M sodium hydroxide for complete neutralisation.Determine the relative atomic mass of M
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