Question

10 ml of solution of acetic acid is added to 250 ml volumetric flask , the concentration of diluted vinegar is 0.01 M. What is the % by mass of acetic acid in vinegar if the density of vinegar is 1.05 g/ml. ( C: 12 O:16 H: 1 )​

Answer 1

The percentage by mass of the acetic acid is 0.057%.

The given parameters:

• Volume of the acetic acid = 10 ml
• Volume of the flask, = 250 ml
• Concentration of the diluted vinegar = 0.01 M
• Density of the vinegar, ρ = 1.05 g/ml

The concentration of the acetic acid is calculated as follows;

$C_1V_1 = C_2 V_2\\\\C_1 = \frac{ C_2 V_2}{V_1} \\\\C_1 = \frac{0.01 \times 250}{10} \\\\C_1 = 0.25 \ M$

The number of moles of acetic acid in the given concentration;

$C = \frac{mol}{L} \\\\mole = CL\\\\mole = 0.25 \times 10 \times 10^{-3} \\\\mole = 0.0025$

The molar mass of acetic acid = 60 g/mol

The mass of acetic acid in the given number of moles is calculated as follows;

m = conc. x molar mass

m = 0.0025 x 60

m = 0.15 g

The mass of vinegar in the volume and density;

m = density x volume

m = 1.05 x 250

m = 262.5 g

The percentage by mass of the acetic acid is calculated as follows;

$= \frac{0.15}{262.5} \times 100\%\\\\= 0.057 \%$

Thus, the percentage by mass of the acetic acid is 0.057%.

Learn more about percentage by mass here: brainly.com/question/22060503