Question

6.025×10^20 molecules of acetic acid are present in 500 ml of its solution.The concentration of solution is

Answer 1

Acetic acid (CH3COOH) is the solute and its 6.025×10^20 molecules are in the solution.

Now, number of moles of the solute (n)=

n = 6.025×10^20/6.022×10^23

n = 10^-3 moles.

And, Volume of the Solution (V) = 500ml

V = 500/1000 = 0.5 letre.

So,

Molarity (M) = n/V

M = 10^-3/0.5

M = 2 × 10^-3 mol/l.....●

Thus, the concentration of the solution is 2×10^-3 mol/l.

Answer 2

Answer:

refer to the above attachment.

Explanation:

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