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A stock solution is made by dissolving 66.05 g of (nh4)2so4 in enough water to make 250 ml of solution. a 10.0 ml sample of this solution is then diluted to 50.0 ml. given that the molar mass of (nh4)2so4 is 132.1 g/mol, what is the concentration of the new solution? use m subscript i v subscript i equals m subscript f v subscript f. and molarity equals startfraction moles of solute over liters of solution endfraction.. 0.400 m 1.60 m 5.00 m 10.0 m

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Eduard22slyidn

The concentration of the new solution of the ammonium sulphate, (NH₄)₂SO₄ given the data is 0.4 M

How to determine the mole of (NH₄)₂SO₄

  • Mass of (NH₄)₂SO₄ = 66.05 g
  • Molar mass of (NH₄)₂SO₄ = 132.1 g/mol
  • Mole of (NH₄)₂SO₄ =?

Mole = mass / molar mass

Mole of (NH₄)₂SO₄ = 66.05 / 132.1

Mole of (NH₄)₂SO₄ = 0.5 mole

How to determine the molarity of (NH₄)₂SO₄

  • Mole of (NH₄)₂SO₄ = 0.5 mole
  • Volume = 250 mL = 250 / 1000 = 0.25 L
  • Molarity of (NH₄)₂SO₄ =?

Molarity = mole / Volume

Molarity of (NH₄)₂SO₄ = 0.5 / 0.25

Molarity of (NH₄)₂SO₄ = 2 M

How to determine the molarity of the diluted solution

  • Volume of stock solution (V₁) = 10 mL
  • Molarity of stock solution (M₁) = 2 M
  • Volume of diluted solution (V₂) = 50 mL
  • Molarity of diluted solution (M₂) =?

M₁V₁ = M₂V₂

2 × 10 = M₂ × 50

20 = V₂ × 50

Divide both side by 50

M₂ = 20 / 50

M₂ = 0.4 M

Learn more about dilution:

https://brainly.com/question/15022582

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Answer:

A: 0.400 M

Explanation:

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