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4 September, 09:08

Rubbing alcohol contains 615g of isopropanol (C3H7OH) per liter (aqueous solution). Calculate the molality of this solution. Given density of rubbing alcohol solution is 0.825 g/cm3 at 22 oC.

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  1. 4 September, 09:23
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    Solution of isopropanol is 10.25 molal

    Explanation:

    615 g of isopropanol (C3H7OH) per liter

    We gave the information that 615 g of solute (isopropanol) are contained in 1L of water. We need to find out the mass of solvent, so we use density.

    Density of water 1g/mL → Density = Mass of water / 1000 mL of water

    Notice we converted the L to mL

    Mass of water = 1000 g (which is the same to say 1kg)

    Molality are the moles of solute in 1kg of solvent, so let's convert the moles of isopropanol → 615 g. 1mol / 60g = 10.25 moles

    Molality (mol/kg) = 10.25 moles / 1kg = 10.25 m
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