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How many moles of aluminum sulfate are required to create 200 L of 3 m solution​

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Answer:

600 moles of (NH₄)₂SO₄ are required

Step-by-step explanation:

If concentration is in m, this means molality.

Molality is reffered to the moles of solute, in 1 kg of solvent.

Volume of solvent = 200L

If we use density, we assume that solvent is water so:

1 g/mL = solvent mass / solvent volume

200L . 1000 = 2x10⁵ mL

1g/mL = solvent mass / 2x10⁵ mL

200000 g = solvent mass

In 1kg of water we have 3 m

In 200 kg of water we have (200 . 3)/1 = 600 moles

User Sanket Vekariya
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