Answer :

Answer : The number of moles of NaCl present in a solution are, 1.07 moles.

Explanation : Given,

Molarity of [tex]NaCl[/tex] = 8.59 M

Volume of solution = 125 mL  = 0.125 L      (1L = 1000 mL)

Molarity : It is defined as the number of moles of solute present in one liter of volume of solution.

Formula used :

[tex]\text{Molarity}=\frac{\text{Moles of }NaCl}{\text{Volume of solution (in L)}}[/tex]

Now put all the given values in this formula, we get:

[tex]8.59M=\frac{\text{Moles of }NaCl}{0.125 L}[/tex]

[tex]\text{Moles of }NaCl=1.07mol[/tex]

Therefore, the number of moles of NaCl present in a solution are, 1.07 moles.

The number of moles of NaCl present in the solution is 1.07 moles.

From the question,

We are to determine the number of moles of NaCl that are present in a solution with a molarity of 8.59 M and a volume of 125mL.

Using the formula

Number of moles = Concentration × Volume

From the given information

Concentration of the solution = 8.59 M

Volume of the solution = 125mL = 0.125 L

∴ Number of moles of NaCl present in the solution = 8.59 × 0.125

Number of moles of NaCl present in the solution = 1.07375 moles

Number of moles of NaCl present in the solution ≅ 1.07 moles

Hence, the number of moles of NaCl present in the solution is 1.07 moles.

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