Determination of the acetic acid (ch3cooh) content of vinegar calculate the molarity of the diluted vinegar using data from one sample (assume the density of vinegar is 1.00 g/ml).

Molarity is the number of moles of solute in one liter of the solution. Mathematically, M = n/V. Where Molarity of the solution is M, Number of moles of solute is n, and volume of the solution is V.

Calculate the mass in 1000 mL of the solution. Density of acetic acid is 1.00 g/mL.

 

Density = 1.00 g/mL. 1.0 mL has mass 1.00 g. 1000.0 mL has mass = (1000.0 mL/ 1.0 mL) * 1.00 g = 1000.0 g. hence, the mass in 1000.0 mL of the solution if 1000.0 g.

 

Calculate the mass contained of 5.00 % acetic acid as follows: 5.00 % of 1040 g = (5.00 / 100) * 1000.0 g = 50.0 g. Hence, the mass contained of 5.00 % acetic acid is 50.0 g.

 

Calculate the number of moles of acetic acid. Number of moles = Mass / Molar mass = 50.0 g / 60.0 g/mol = 0.8333 mol. Hence, the number of moles of acetic acid is 0.8333 mol.

 

Calculate the molarity of the solution, M = n / V = 0.8333 mol / 1.0 L = 0.8333 M

 

Therefore, the molarity of the solution if 0.8333 M

 

The answer is tea due to the main ingredient being a herb that grows from the ground and sugar that grows from a stalk, i hope this : )     p.s.  can i be brainliest?

Ithink the answer is b.