Question

A student wants to prepare a solution of acetic acid and water
having 30% v/v. If 6 mL of acetic acid has
been taken then how
much water must be added to get the desired concentration?​

Answer 1

Given:-

→ Volume of acetic acid = 6 mL

→ Concentration [v/v %] = 30%

To find:-

→ Volume of water.

Solution:-

Let the required volume of water be

'x' mL.

Here, acetic acid is solute and water is solvent. We are given with the volume of solute and concentration of solution. So firstly, let's calculate the total volume of the solution.

Volume of solution :-

= Volume of solute + Volume of solvent

= 6 mL + x mL

= (6 + x) mL

Concentration of a solution [v/v %] :-

= Vol. of solute/Vol. of solution × 100

By substituting values, we get :-

=> 6/(6 + x) × 100 = 30

=> 6 × 100 = 30(6 + x)

=> 600 = 180 + 30x

=> 30x = 600 - 180

=> 30x = 420

=> x = 420/30

=> x = 14 mL

Thus, 14 mL of water must be added to get the desired concentration.

Answer 2

We have,

v/v% concentration of solution = 30%

Volume of Acetic acid = 6 mL

Let, volume of water required = (x) mL

∴ Volume of Solution = (6 + x) mL

We know,

v/v% of concentration = (Volume of solute)/(Volume of solution) × 100%

∴ (6)/(6 + x) × 100 = 30

⇒ 30x + 1800 = 600

⇒ 30x = 600 - 180 = 420

⇒ 3x = 42

⇒ x = 14 mL {Answer}.