G In Part 7, the [Cl-] in saturated NaCl is 5.4 M at room temperature. Assume that you had 1.00 ml of the saturatedsolution, and that you added 0.50 ml of 12 M HCl. What is the [Cl-] after you added the HCl. (When two solutionscontain the same component, the numerator consists of the sum of the volume times the concentration for each solu-tion. The denominator is the total volume.

7.60 M

Explanation:

Our method to solve this question is to  use the definition of molarity (M) concentration which is the number of moles per liter of solution, so for this problem we have

[Cl⁻] = # mol Cl⁻ / Vol

Now the number of moles of Cl⁻ will be sum of Cl in the 1.00 mL 5.4 M solution plus the moles of Cl⁻ in the 0.50 mL 12 M H . Since the volume in liters times the molarity gives us the number of moles we will have previous conversion of volume to liters for units consistency:

1mL x 1 L / 1000 mL = 0.001 L

0.5 mL x 1L/1000 mL = 0.0005 L

[Cl⁻]  =  0.001 L x 5.4 mol/L + 0.0005L x 12 mol/L / ( 0.001 L + 00005 L )

= 7.6 M

This is the same as the statement given in the question.

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Explanation: