Question

6.025 x 10^20 molecules of acetic acid are present in 500 ml of its solution the concentration of solution is

Answer 1

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Answer 2

$\begin{array}{l}{6.022 \times 10^{23} \text { molecules }=1 \text { mole }} \\ {6.022 \times 10^{20} \text { molecules }=\left(1 \text { mole } / 6.022 \times 10^{23} \text { molecules) }\right.} \\ {6.022 \times 10^{20} \text { molecules }=0.001 \text { moles }}\end{array}$

Given:  

Volume = 500 mL  =  0.500 L

Concentration = Moles / volume (L)

Plug the given values in the formula:

Concentration = 0.001 moles / 0.500 L  

Concentration = 0.002 M  

Hence $6.025 \times 10^{20}$ molecules of ‘acetic acid’ are present in 500 ml of its solution the ‘concentration of solution’ is 0.002 M.