A solution is 5.5% (w/v) ethanol (C2H5OH, MW 46.07 g/mol). What is the molar concentration of ethanol in this solution?
DrBob, Correct me please, but wouldn't 5.5% w/v => 5.5-g EtOH/100-ml soln (assuming water solvent)? => (5.5-g/46.07-g/mol) / (0.100-L) = 1.2M in EtOH (2 sig. figs.)
w/v of what solvent ( perhaps water).What is the density of the solution?
To find the molar concentration of ethanol in the solution, you need to convert the given mass/volume percentage concentration (w/v) to molar concentration.
First, let's understand the given information:
- Solution: 5.5% (w/v) ethanol
- Molecular weight (MW) of ethanol (C2H5OH): 46.07 g/mol
The mass/volume percentage concentration (w/v) represents the mass of the solute (ethanol) per unit volume of the solution, expressed as a percentage.
So, for every 100 mL (or 100 g) of the solution, there are 5.5 g of ethanol.
To determine the molar concentration, we need to convert the mass of ethanol to moles using its molecular weight and then divide it by the volume of the solution.
Step 1: Convert the mass of ethanol to moles.
Mass of ethanol in the solution = 5.5 g
Moles of ethanol = mass / molecular weight
= 5.5 g / 46.07 g/mol
≈ 0.1193 mol
Step 2: Determine the molar concentration.
Molar concentration = moles of solute / volume of solution (in liters)
≈ 0.1193 mol / 1 L
≈ 0.1193 M
Therefore, the molar concentration of ethanol in the solution is approximately 0.1193 M.