A is a solution containing 0.05 mol/dm-3 of H2SO4.B is a solution containing 1.4g per 25 cm3 of XOH.25.0cm3 of B neutralized 24.50cm3 of A on titration using methyl orange as indicator calculate the following.(1)concentration of B in mol/dm-3.(2)molar mass of B. (3)Relative atomic mass of X in XOH.(4)Percentage by mass of X inXOH.
(1) Let's start by calculating the concentration of solution B in mol/dm³.
We know that 25.0 cm³ of B neutralized 24.50 cm³ of A. Since methyl orange is used as the indicator, we assume that the number of moles of H₂SO₄ in A equals the number of moles of XOH in B.
We have:
Number of moles of H₂SO₄ = 0.05 mol/dm³ × 0.02450 dm³ = 0.001225 moles
Now, we need to find the number of moles of XOH in B.
First, let's calculate the mass of XOH in 25.0 cm³ of B:
Mass of XOH = 1.4 g
Next, convert the mass of XOH to moles using its molar mass:
Molar mass of XOH = Mass/Volume
Molar mass of XOH = (1.4 g) / (0.0250 dm³) = 56 g/dm³
Now, let's calculate the number of moles of XOH:
Number of moles of XOH = Mass/Molar mass
Number of moles of XOH = (1.4 g) / (56 g/dm³) = 0.025 moles
Since the moles of H₂SO₄ and XOH are equal (according to the titration), the concentration of solution B is:
Concentration of B = Number of moles/Volume
Concentration of B = 0.025 moles / 0.025 dm³ = 1.0 mol/dm³
(2) The molar mass of B can be calculated as follows:
Molar mass of XOH = Mass/Number of moles
Molar mass of XOH = (1.4 g) / (0.025 moles) = 56 g/mol
(3) The relative atomic mass of X in XOH can be determined from the molar mass of XOH above:
Relative atomic mass of X = Molar mass of XOH - Molar mass of OH
Relative atomic mass of X = 56 g/mol - 17 g/mol (Molar mass of OH) = 39 g/mol
(4) The percentage by mass of X in XOH can be calculated as follows:
Percentage by mass of X in XOH = (Molar mass of X / Molar mass of XOH) × 100
Percentage by mass of X in XOH = (39 g/mol / 56 g/mol) × 100 ≈ 69.64% (rounded to two decimal places)
So, the percentage by mass of X in XOH is approximately 69.64%.