PLEASE SOMEONE! The density of the acetic acid solution is 1.05 g/mL. Calculate the %(m/m) of the acetic acid solution (convert mol/L to %(m/m)). CH3COOH(aq) + NaOH(aq) = CH3COOHNa(aq) + H2O(L)
Volume of CH3COOH = 0.1L
Titration Equation: CH3COOH(aq) + NaOH(aq) = CH3COOHNa(aq) + H2O(L)
Concentration of CH3COOH = 0.243 mol/L

The Percentage concentration of acetic acid = 1.39 %

Explanation:

Density of acetic acid solution = 1.05 g/mL

Volume of acetic acid solution = 0.1 L = 100 mL

From the formula, Density = mass / volume; mass = density × volume

Mass of acetic acid solution = 1.05 g/mL × 100 mL = 105 g

Molar concentration of acetic acid solution = 0.243 mol/L

Molar mass of acetic acid, CH₃COOH = (12 × 2 + 1 ×4 + 16 ×2) = 60 g/mol

From the formula, mass concentration = molar concentration × molar mass

Mass concentration of acetic acid, CH₃COOH = 0.243 mol/L × 60 g/mol = 14.58 g/L

In one liter of acetic acid solution, there are 14.58 g of acetic acid. Therefore, in 0.1 L, there will be 14.58 × 0.1 = 1.458 g of acetic acid.

Percentage concentration of acetic acid = mass of acetic acid / mass of acetic acid solution × 100%

Percentage concentration of acetic acid = (1.458 / 105) × 100% = 1.39 %

The Percentage concentration of acetic acid = 1.39 %

the answer is c

tell me if wrong