A 15.5 mL sample of 0.215 M KOH solution required 21.2 mL of aqueous acetic acid solution in a titration experiment. Calculate the molarity of the acetic acid solution.

0.156 M

Explanation:

The computation of the molarity of the acetic acid solution is shown below:

The balanced chemical reaction is

As we know that The KOH and acetic acid would have 1:1 ratio. Here we need to determine the moles of KOH

So,

moles KOH = 0.0155 L ×0.214 mol/L

= 3.317x10-3 mol

Now moles CH_3COOH = 3.317x10-3 mol

[CH3COOH] = moles ÷ volume

= 3.317 × 10^-3 mol ÷ 0.0212 L

= 0.156 M

a  solution  is always transparent, light passes through with no scattering from solute particles which are molecule in size. the  solution  is homogeneous and does not settle out. a colloid is intermediate between asolution  and a  suspension. while asuspension  will separate out a colloid will not.