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What is the molartiy of so4 -2 ion in aqueous solution that contain 34.2 ppm of Al2(SO4)3 ? (Assume complete dissociation and density of solution 1 g/mL) (a) 3 x 10-4 M (b) 2 x 104 M (C) 104 M (d) None of these?
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What is the molartiy of so4 -2 ion in aqueous solution that contain 34...
Calculation of Molarity of SO4 -2 Ion in Aqueous Solution
Given:
- Concentration of Al2(SO4)3 solution = 34.2 ppm
- Density of solution = 1 g/mL
To calculate the molarity of SO4 -2 ion in the solution, we need to follow the following steps:
Step 1: Calculate the molar mass of Al2(SO4)3
Al2(SO4)3 = (2 × 27) + (3 × 32 + 4 × 16) = 342 g/mol
Step 2: Calculate the number of moles of Al2(SO4)3 present in 1 L of the solution
1 L of solution = 1000 mL
Concentration of Al2(SO4)3 solution = 34.2 ppm
Therefore, 34.2 mg of Al2(SO4)3 is present in 1 L of the solution
Number of moles of Al2(SO4)3 = (34.2 mg / 342 g/mol) × (1 mol / 1000 mg) = 1 × 10^-4 mol
Step 3: Calculate the number of moles of SO4 -2 ion present in 1 L of the solution
1 mole of Al2(SO4)3 contains 3 moles of SO4 -2 ions
Therefore, number of moles of SO4 -2 ion = 3 × 1 × 10^-4 mol = 3 × 10^-4 mol
Step 4: Calculate the molarity of SO4 -2 ion in the solution
Molarity = (number of moles of solute) / (volume of solution in L)
Volume of solution = Density of solution × Mass of solution
Mass of solution = Concentration of Al2(SO4)3 solution × Volume of solution
= 34.2 mg/L × 1000 mL / 1000 L = 34.2 g/L
Volume of solution = 34.2 g/L / 1 g/mL = 34.2 mL
Molarity of SO4 -2 ion = (3 × 10^-4 mol) / (34.2 mL / 1000 mL/L) = 8.77 × 10^-3 M
Therefore, the molarity of SO4 -2 ion in the solution is 8.77 × 10^-3 M.
Answer: None of these.
Given:
- Concentration of Al2(SO4)3 solution = 34.2 ppm
- Density of solution = 1 g/mL
To calculate the molarity of SO4 -2 ion in the solution, we need to follow the following steps:
Step 1: Calculate the molar mass of Al2(SO4)3
Al2(SO4)3 = (2 × 27) + (3 × 32 + 4 × 16) = 342 g/mol
Step 2: Calculate the number of moles of Al2(SO4)3 present in 1 L of the solution
1 L of solution = 1000 mL
Concentration of Al2(SO4)3 solution = 34.2 ppm
Therefore, 34.2 mg of Al2(SO4)3 is present in 1 L of the solution
Number of moles of Al2(SO4)3 = (34.2 mg / 342 g/mol) × (1 mol / 1000 mg) = 1 × 10^-4 mol
Step 3: Calculate the number of moles of SO4 -2 ion present in 1 L of the solution
1 mole of Al2(SO4)3 contains 3 moles of SO4 -2 ions
Therefore, number of moles of SO4 -2 ion = 3 × 1 × 10^-4 mol = 3 × 10^-4 mol
Step 4: Calculate the molarity of SO4 -2 ion in the solution
Molarity = (number of moles of solute) / (volume of solution in L)
Volume of solution = Density of solution × Mass of solution
Mass of solution = Concentration of Al2(SO4)3 solution × Volume of solution
= 34.2 mg/L × 1000 mL / 1000 L = 34.2 g/L
Volume of solution = 34.2 g/L / 1 g/mL = 34.2 mL
Molarity of SO4 -2 ion = (3 × 10^-4 mol) / (34.2 mL / 1000 mL/L) = 8.77 × 10^-3 M
Therefore, the molarity of SO4 -2 ion in the solution is 8.77 × 10^-3 M.
Answer: None of these.
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What is the molartiy of so4 -2 ion in aqueous solution that contain 34...
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What is the molartiy of so4 -2 ion in aqueous solution that contain 34.2 ppm of Al2(SO4)3 ? (Assume complete dissociation and density of solution 1 g/mL) (a) 3 x 10-4 M (b) 2 x 104 M (C) 104 M (d) None of these?
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What is the molartiy of so4 -2 ion in aqueous solution that contain 34.2 ppm of Al2(SO4)3 ? (Assume complete dissociation and density of solution 1 g/mL) (a) 3 x 10-4 M (b) 2 x 104 M (C) 104 M (d) None of these? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about What is the molartiy of so4 -2 ion in aqueous solution that contain 34.2 ppm of Al2(SO4)3 ? (Assume complete dissociation and density of solution 1 g/mL) (a) 3 x 10-4 M (b) 2 x 104 M (C) 104 M (d) None of these? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for What is the molartiy of so4 -2 ion in aqueous solution that contain 34.2 ppm of Al2(SO4)3 ? (Assume complete dissociation and density of solution 1 g/mL) (a) 3 x 10-4 M (b) 2 x 104 M (C) 104 M (d) None of these?.
What is the molartiy of so4 -2 ion in aqueous solution that contain 34.2 ppm of Al2(SO4)3 ? (Assume complete dissociation and density of solution 1 g/mL) (a) 3 x 10-4 M (b) 2 x 104 M (C) 104 M (d) None of these? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about What is the molartiy of so4 -2 ion in aqueous solution that contain 34.2 ppm of Al2(SO4)3 ? (Assume complete dissociation and density of solution 1 g/mL) (a) 3 x 10-4 M (b) 2 x 104 M (C) 104 M (d) None of these? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for What is the molartiy of so4 -2 ion in aqueous solution that contain 34.2 ppm of Al2(SO4)3 ? (Assume complete dissociation and density of solution 1 g/mL) (a) 3 x 10-4 M (b) 2 x 104 M (C) 104 M (d) None of these?.
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