Calculate the molality of a solution formed by adding 3.331 g of methanol (CH3OH) to 100.0 mL of water. The density of water is 0.9982 g/mL.
Answer:
Given the mass of methanol=3.331 g,
and the molar mass of methanol=32.04 g/mol.
Therefore moles of methanol=mass/molar mass=3.331 g/32.04 g/mol=0.10396 mol.
Given the volume of water=100 mL and the density =0.9982 g/mL
Mass of water=density x volume=0.9982 g/mL x 100 mL=99.82 g=0.09982 Kg.
Molality=moles of the solute/mass of solvent(Kg)
Here solute is methanol and solvent is water.
Molality=0.10396 mol/0.09982 Kg=1.0415 mol/Kg
Molality=1.0415 m.
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Calculate the molality of a solution formed by adding 3.331 g of methanol (CH3OH) to 100.0...
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