How many mL of a 12.5% (by volume) solution of ethanol are required to produce 175 mL of 0.664 M acetic acid? (density of pure ethanol = 0.789 g/mL)
We have the following data:
% m1/m (percent by solute mass/solution mass) = 12.5
Molar Mass (MM) of ethanol (C2H5OH) ≈ 46.07 g/mol
M1 (molarity) = 0.664 M (or 0.664 mol/L)
V1 (initial volume) = 175 ml = 0.175 L
n (number of mol) = ?
m1 (solute mass - ethanol) = ?
m (solution mass) = ?
d (density of pure ethanol) = 0.789 g/ml
v (volume of pure ethanol) = ? (in ml)
[First Step] Let's find the number of moles or matter number, let's see:
[Second Step] We apply the data to the Molarity or Molar Concentration formula and find the mass of the solute (m1), let's see:
[Third Step] We apply the data to the formula of percent solute mass per solution mass and let's find the solution mass, let's see:
[Fourth Step] We apply the data to the density formula and let's find the volume of ethanol, let's see:
The volume is approximately 54.3 ml
Atomic Number is "8"
Molarity of solution: 1.1884 MFurther explanation
The concentration of a substance can be expressed in several quantities such as moles, percent (%) weight / volume,), molarity, molality, parts per million (ppm) or mole fraction. The concentration shows the amount of solute in a unit of the amount of solvent.
Molarity shows the number of moles of solute in every 1 liter of solution or mmol in each ml of solution
M = Molarity
n = Number of moles of solute
V = Volume of solution
Mole itself is the number of particles contained in a substance
1 mole = 6.02.10²³ particles
Mole can be obtained if the amount of substance mass and its molar mass is known
KOH (Potassium hydroxide)
Molar mass: Ar K + Ar O + Ar H
molar mass: 39,0983 + 15,999 + 1,008
molar mass: 56.1053 g / mol
15 g of KOH
225 ml of solution = 0.225 LAsked
The molar concentration of a solutionSolution
Molarity (M) of a solution
atoms in 5.1 moles of Sulfur
the mole fractions of the solute and the solvent
moles of CaCO3
the mole ratio of Al to Cl2