An aqueous solution with a density of 0.998 g/ml at 20 degrees Celsius is prepared by dissolving 12.8 mL of CH3CH2CH2OH (p=0.803 g/mL) in enough water to make 75.0 mL of solution calculate the percent (w/w), percent (w/v), parts per million, and molality?

Question

An aqueous solution with a density of 0.998 g/ml at 20 degrees Celsius is prepared by dissolving 12.8 mL of CH3CH2CH2OH (p=0.803 g/mL) in enough water to make 75.0 mL of solution calculate the percent (w/w), percent (w/v), parts per million, and molality?

Answer 1

To calculate the percent (w/w), percent (w/v), parts per million (ppm), and molality of the aqueous solution, we need to determine the mass of solute and solvent present in the solution.

Let's first calculate the mass of the solute (CH3CH2CH2OH):

Mass of CH3CH2CH2OH = volume x density = 12.8 mL x 0.803 g/mL = 10.3024 g

Now, let's calculate the mass of the solvent (water) present in the solution:

Volume of water = total volume of solution - volume of solute
Volume of water = 75.0 mL - 12.8 mL = 62.2 mL (Note: We assume that the volume of the solute does not change upon dissolution in water.)

To calculate the percent (w/w):

percent (w/w) = (mass of solute / mass of solution) x 100

mass of solution = mass of solute + mass of solvent

mass of solution = 10.3024 g + mass of water

mass of water = volume of water x density of water = 62.2 mL x 0.998 g/mL = 61.9728 g

mass of solution = 10.3024 g + 61.9728 g = 72.2752 g

percent (w/w) = (10.3024 g / 72.2752 g) x 100 = 14.26% (rounded to two decimal places)

To calculate the percent (w/v):

percent (w/v) = (mass of solute / volume of solution) x 100

percent (w/v) = (10.3024 g / 75.0 mL) x 100 = 13.74% (rounded to two decimal places)

To calculate the parts per million (ppm):

ppm = (mass of solute / mass of solution) x 10^6

ppm = (10.3024 g / 72.2752 g) x 10^6 = 142,665 ppm (rounded to the nearest whole number)

To calculate the molality:

Molality (m) = (moles of solute / mass of solvent in kg)

First, we need to calculate the number of moles of solute:

moles of solute = mass of solute / molar mass of CH3CH2CH2OH

The molar mass of CH3CH2CH2OH (propanol) is:

C: 12.01 g/mol
H: 1.008 g/mol (x 3)
O: 16.00 g/mol

Molar mass of CH3CH2CH2OH = (12.01 g/mol) + (1.008 g/mol x 3) + (16.00 g/mol) = 60.09 g/mol

moles of solute = 10.3024 g / 60.09 g/mol = 0.1712 mol (rounded to four decimal places)

Next, we need to convert the mass of the solvent from grams to kilograms:

mass of solvent = mass of water / 1000

mass of solvent = 61.9728 g / 1000 = 0.0619728 kg

Molality (m) = (0.1712 mol / 0.0619728 kg) = 2.7623 mol/kg (rounded to four decimal places)

So, the calculations are as follows:

- Percent (w/w): 14.26%
- Percent (w/v): 13.74%
- Parts per million (ppm): 142,665 ppm
- Molality: 2.7623 mol/kg