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 molarity
grams CH3CH2CH2OH = volume x density = 12.8 x 0.803 = ?
moles CH3CH2CH2OH = grams/molar mass
Then M = moles/L soln.
L soln = 0.075L
To calculate the molarity (M) of the solution, we need to know the moles of the solute (CH3CH2CH2OH) and the volume of the solution in liters.
First, let's calculate the moles of CH3CH2CH2OH using its density and volume.
Mass of CH3CH2CH2OH = density × volume
= 0.803 g/mL × 12.8 mL
= 10.2784 g
Next, we need to convert the mass of CH3CH2CH2OH to moles using its molecular weight. The molecular weight of CH3CH2CH2OH (propanol) is 60.11 g/mol.
Moles of CH3CH2CH2OH = mass / molecular weight
= 10.2784 g / 60.11 g/mol
= 0.171 moles
Now, we need to calculate the volume of the solution in liters. We are given that the density of the solution is 0.998 g/mL and the volume of the solution is 75.0 mL.
Volume of solution = volume of water + volume of CH3CH2CH2OH
= 75.0 mL
Since the volume of water is not provided, we will assume that the volume of CH3CH2CH2OH remains constant when dissolved in water.
Volume of CH3CH2CH2OH = 12.8 mL
Volume of water = Volume of solution - Volume of CH3CH2CH2OH
= 75.0 mL - 12.8 mL
= 62.2 mL
Now, we convert the volume of water to liters.
Volume of water = 62.2 mL * (1 L / 1000 mL)
= 0.0622 L
Finally, we can calculate the molarity by dividing the moles of CH3CH2CH2OH by the volume of the solution in liters.
Molarity (M) = moles / volume (in liters)
= 0.171 moles / 0.0622 L
≈ 2.75 M
Therefore, the molarity of the solution is approximately 2.75 M.