Question

5% acetic acid in water. This is a volume percent (v/v) which means there are 5 4. Vinegar is a solution of approximately of acetic acid per 100 mL. of vinegar. The density of acetic acid is 1.05 g/mL. Use stoichiometry to determine how man grams of acetic acid are in 15.0 mL of vinegar. are the molar masses of the two reactants, acetic acid and sodium bicarbonate? Show all work

Answer 1

4 Ans.

The vinegar solution has 5% (v/v) acetic acid. That means, there are 5 mL of acetic acid in 100 mL vinegar solution.

In 15.0 mL solution of vinegar, 5% volume of it is acetic acid.

Therefore, volume of acetic acid is

$= \frac{5}{100}\times 15.0mL$

$= 0.75 mL$

Therefore, 0.75 mL is acetic acid.

Given that, density of acetic acid is 1.05g/mL. It means, every mL of acetic acid weighs 1.05g.

Therefore, amount of acetic acid in 0.75mL acetic acid is

$= d \times V$

$= 1.05 g/mL \times 0.75 mL$

$= 0.788g$

Therefore, there are 0.788g of acetic acid in 15.0 mL of vinegar solution.

5 Ans.

Molar mass can be calculated by adding the atomic mass of each of the atoms in the compound. If two atoms of the same element is present, then the atomic mass is multiplied by 2 and added to the other atomic masses.

Molar mass od acetic acid, CH₃COOH is

$= (12\times 2)+ (1\times 4)+ (16\times 2)$

There are 2 atoms of C, 4 atoms of H and 2 atoms of O.

$= 24 + 4 + 32$

$= 60g/mol$

Therefore, the molar mass of acetic acid is 60g/mol.

Molar mass of sodium bicabonate, NaHCO₃ is

$= 23 + 1 + 12 + (16\times 3)$

$= 23 + 1 + 12 + 48$

$= 84g/mol$

Therefore, the molar mass of sodium bicarbonate is 84g/mol.