Question

A solution contains 1.86 g of methaneboronic acid, CBH5O2, dissolved in 77.7 mL of acetone (p = 0.818 g/mL). Calculate the molality, mole fraction, and mass percent of methaneboronic acid in the solution.

Answer 1

acetone 77.7 ml with a density of 0.818 g/ml

mass of acetone = density * volume

mass = 77.7 * 0.818 = 63.5586 grams

molar mass of acetone is 58 g/gmol

moles = mass / molar mass

moles of acetone = 63.55 / 58 = 1.0956 moles of acetone

moles of methaneboronoc acid

I´m getting a value of 60 g/gmol

moles of acid is 1.86 / 60 = 0.031moles

mole fraction = moles of acid / total moles

mole fraction = 0.031 / (0.031 + 1.0956) = 0.02751 , multiply this by 100 to get

mole fraction = 2.75 %

mass percent = mass of acid / total mass

= 1.86 / (1.86 + 63.55) = 0.0284 multiply this by 100 to get 2.84%

molality is moles of solute / kg of solvent

moles of acid = 0.031

mass of solvent = 63.5586 grams, divide by 1000 to get 0.06355 kg

molality = 0.031 / 0.06355 = 0.487