EXPERIMENT 2: Preparation of ca 0.1 mol.dm3 HCI SolutionPre-calculation:mHCl solution = 1,16 x 103 x 32 = 3,712 x 10² g100CHCI-soluion = 3,712 x 10236,46 x 1dm3= 10,18 mol.dm 3CV1 = C2V20,1 mol.dm-3 x 500 mL10,18 mol.dm3VHC-Solution= C2V2 == 4,9 mLTask:Calculate the volume of the sameto 1000 cm to make a 0.05 mol.dm3 HCl solution at room temperature.centrated HCl as from the above expiment which can be diluted

Answered: Image /qna-images/answer/11fb3d1d-be28-4c4a-957c-143c70cc14ce.jpg

EXPERIMENT 2: Preparation of ca 0.1 mol.dm HCI Solution Pre-calculation: mHC-solaton = 1,16 x 103 x 32 = 3,712 x 102 g 100 !3! CHCL = 3,712 x 102 36,46 x 1dm3 %3D = 10,18 mol.dm 3 CV = C2V2 0,1 mol.dm3 x 500 mL 10,18 mol.dm3 %3D HCI-solution = 4,9 mL %3D C1 Task: Calculate the volume of the same concentrated HCl as from the above experiment which can be diluted to 1000 cm to make a 0.05 mol.dm HCl solution at room temperature.

EXPERIMENT 2: Preparation of ca 0.1 mol.dm HCI Solution Pre-calculation: mHC-solaton = 1,16 x 103 x 32 = 3,712 x 102 g 100 !3! CHCL = 3,712 x 102 36,46 x 1dm3 %3D = 10,18 mol.dm 3 CV = C2V2 0,1 mol.dm3 x 500 mL 10,18 mol.dm3 %3D HCI-solution = 4,9 mL %3D C1 Task: Calculate the volume of the same concentrated HCl as from the above experiment which can be diluted to 1000 cm to make a 0.05 mol.dm HCl solution at room temperature.

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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