Chemistry help molality?

Perspiration has a slight acidity due to the presence of lactic acid (C3H6O3, MW = 90.08 g/mol). Suppose we analyze the perspiration of a chemistry student running late to class. If the density of the perspiration was 1.15 g/mL and the mass percent of lactic acid was found to be 4.88%, what would be the molality of the lactic acid in the perspiration?
The answer is 0.570 m but I cannot reach this... Can someone show me the steps to do this?

1 Answer
Ernest Z.
Feb 6, 2018

#b = "0.570 mol/kg"#

Explanation:

Assume that the volume of perspiration is 100 mL.

Step 1. Calculate the mass of the perspiration

#"Mass" = 100 color(red)(cancel(color(black)("mL"))) × "1.15 g"/(1 color(red)(cancel(color(black)("mL")))) = "115 g"#

Step 2. Calculate the masses of lactic acid and water

#"Mass of lactic acid" = 115 color(red)(cancel(color(black)("g perspiration"))) × "4.88 g lactic acid"/(100 color(red)(cancel(color(black)("g perspiration")))) = "5.612 g lactic acid"#

#"Mass of water = (115 - 5.612) g = 109.388 g"#

Step 3. Calculate the moles of lactic acid

#"Moles of lactic acid" = 5.612 color(red)(cancel(color(black)("g lactic acid"))) × "1 mol lactic acid"/(90.08 color(red)(cancel(color(black)("g lactic acid")))) = " 0.062 30 mol lactic acid"#

Step 4. Calculate the molal concentration of lactic acid

#b = "moles of solute"/"kilograms of solvent" = "0.062 30 mol"/"0.109 388 kg" = "0.570 mol/kg"#

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