Answered: How much CaCl2 would be required to…

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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How much CaCl₂ would be required to make 50.0mL of a solution with 0.75 M Cl^- ion? (Use 1.00g/mL as the density for solution and water as the full volume of solution)

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$M o l a r i t y = \frac{N o . o f m o l e s o f C l -}{V o l o f s o l^{n} (m L)} \times 100 0.75 = \frac{n}{50.0 m^{2}} \times 100 n = \frac{0.75 \times 50}{1000} n = 0.0375 m o l e s o f C l^{-} N o w, C a C l_{2} \to C a^{2 +} + 2 C l^{-} 1 m o l e o f C a C l_{2} g i v e s 2 m o l e s o f C l^{-} 2 m o l e s o f C l^{-} f o r m e d = 1 m o l e o f C a C l^{-} 0.0375 m o l e s o f C l^{-} f o r m e d = \frac{1}{2} \times 0.0375 = 0.01875 m o l e s M o l e c u l a r M a s s o f C a C l_{2} = 110.98 \frac{g}{m o l} M a s s o f C a C l_{2} n e e d e d = m o l e s o f C a C l_{2} \times M o l e c u l a r M a s s = 110.98 \times 0.01875 = 2.08 g$

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