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A salt mixture (1.0gm) contains 25%wt of mgso4 and 75% of M2so4.Aqueous solution of this salt mixture on treating with excess bacl2 solution results in the precipitation of of 1.49g of baso4.The atomic mass of M in g/mol is?
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A salt mixture (1.0gm) contains 25%wt of mgso4 and 75% of M2so4.Aqueou...
Calculating the moles of BaSO4 Precipitated

  • Mass of BaSO4 precipitated = 1.49g

  • Molar mass of BaSO4 = 233.38 g/mol

  • Moles of BaSO4 precipitated = Mass/Molar mass = 1.49/233.38 = 0.00639 mol



Calculating the moles of MgSO4 in the salt mixture

  • Percentage of MgSO4 in the salt mixture = 25%

  • Mass of the salt mixture = 1.0g

  • Mass of MgSO4 in the salt mixture = Percentage/100 x mass of salt mixture = 0.25 x 1.0 = 0.25g

  • Molar mass of MgSO4 = 120.37 g/mol

  • Moles of MgSO4 in the salt mixture = Mass/Molar mass = 0.25/120.37 = 0.00208 mol



Calculating the moles of M2SO4 in the salt mixture

  • Percentage of M2SO4 in the salt mixture = 75%

  • Mass of the salt mixture = 1.0g

  • Mass of M2SO4 in the salt mixture = Percentage/100 x mass of salt mixture = 0.75 x 1.0 = 0.75g

  • Molar mass of M2SO4 = 2 x Molar mass of M + Molar mass of S + 4 x Molar mass of O

  • Molar mass of M2SO4 = 2M + 32 + 64 = 2M + 96 g/mol

  • Moles of M2SO4 in the salt mixture = Mass/Molar mass = 0.75/(2M + 96)



Calculating the atomic mass of M

  • The reaction between BaCl2 and the salt mixture can be represented as:

  • BaCl2 + M2SO4 → BaSO4 + 2MCl

  • The balanced equation shows that 1 mole of M2SO4 reacts with 1 mole of BaCl2 to produce 1 mole of BaSO4

  • Therefore, moles of M2SO4 = moles of BaSO4 = 0.00639 mol

  • Substituting the values of moles of M2SO4 in the equation Moles of M2SO4 = Mass/Molar mass, we get:

  • 0.00639 = 0.75/(2M + 96)

  • Solving for M, we get M = 24.31 g/mol



Therefore, the atomic mass of M in the salt mixture is 24.31 g/mol.
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A salt mixture (1.0gm) contains 25%wt of mgso4 and 75% of M2so4.Aqueous solution of this salt mixture on treating with excess bacl2 solution results in the precipitation of of 1.49g of baso4.The atomic mass of M in g/mol is?
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A salt mixture (1.0gm) contains 25%wt of mgso4 and 75% of M2so4.Aqueous solution of this salt mixture on treating with excess bacl2 solution results in the precipitation of of 1.49g of baso4.The atomic mass of M in g/mol is? for Chemistry 2024 is part of Chemistry preparation. The Question and answers have been prepared according to the Chemistry exam syllabus. Information about A salt mixture (1.0gm) contains 25%wt of mgso4 and 75% of M2so4.Aqueous solution of this salt mixture on treating with excess bacl2 solution results in the precipitation of of 1.49g of baso4.The atomic mass of M in g/mol is? covers all topics & solutions for Chemistry 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A salt mixture (1.0gm) contains 25%wt of mgso4 and 75% of M2so4.Aqueous solution of this salt mixture on treating with excess bacl2 solution results in the precipitation of of 1.49g of baso4.The atomic mass of M in g/mol is?.
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