**Problem Statement:**How many mL of 0.500 M NaI would be required to make a 0.0715 M solution of NaI when diluted to 275.0 mL with water?**Explanation:**This problem involves diluting a more concentrated solution to obtain a less concentrated one. The relationship between the concentrations and volumes of the solutions before and after dilution can be described with the equation:\[ C_1 \times V_1 = C_2 \times V_2 \]Where:- \( C_1 \) = initial concentration (0.500 M)- \( V_1 \) = volume of the initial concentration solution (unknown)- \( C_2 \) = final concentration (0.0715 M)- \( V_2 \) = final total volume after dilution (275.0 mL)To find \( V_1 \), rearrange the equation:\[ V_1 = \frac{C_2 \times V_2}{C_1} \] Substitute the given values:\[ V_1 = \frac{0.0715 \, \text{M} \times 275.0 \, \text{mL}}{0.500 \, \text{M}} \] Calculate \( V_1 \) to determine the required volume of the 0.500 M NaI solution.

In the dilution process, the relationship between initial and final concentrations and volumes of solutions are given in the volumetric equation: McVc=MdVdMc-Initial concentrationMd-Final concentrationVc-Initial volumeVd-Final volumeInitial volume can be calculated as shown below: McVc=MdVdVc=MdVdMcVc=0.0715 M ×275.0 mL0.500 MVc=39.3 mL

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How many mL of 0.500 M Nal would be required to make a 0.0715 M solution of Nal when diluted to 275.0 mL with water?

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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