12 13 4 Farmingdale State College CHM 380 Laboratory Maal Name: D: Lab #2: Pre-L

Question

12 13 4 Farmingdale State College CHM 380 Laboratory Maal Name: D: Lab #2: Pre-Lab Assignment Complete in advance and turn in at the beginning of class. The following ng questions require the use of a dilution equation (for example CiV CVa). Please read the introductory material for the lab for a refresher on this topic. 1. Calculate the volume of 2.00 M NaOH required to prepare 100.0 mL of a 0.300 M solution of NaOH. 2. A 50.0 mL sample of 0.26 M Na Acetate is diluted to 780 mL. What is the molarity of the resulting solution? 3. How many milliters of water must be added to a 120.0 mL of 1.80 M HCI to give 1.00 M HCI? (NOTE: You need to calculate final volume FIRST) 4. To 10.0 mL of a 0.70 M CaClh solution is added 25 mlL of water. What is the final concentration? (NOTE: What is the final volume?) 15

Explanation / Answer

Ans. #1. Using C1V1 (original soln.) = C2V2 (Final soln.)

            V1 = (0.300 M x 100.0 mL) / 2.00 M

            Hence, V1 = 15.0 mL

Therefore, required vol. of 2.00 M NaOH (original soln.) = 15.0 mL

#2. Using C1V1 (original soln.) = C2V2 (Final soln.)

            M2 = (0.26 M M x 50.0 mL) / 780.0 mL

            Hence, M2 = 0.0167 M

#3. Using C1V1 (original soln.) = C2V2 (Final soln.)

            V2 = (1.80 M x 120.0 mL) / 1.00 M = 216.0 mL

Therefore, total volume of final solution = 216.0 mL

Now,

            Required vol. of water = V2 – V1 = 216.0 mL – 120.0 mL = 96.0 mL

#4. Total volume of final solution = V1 + V2 = 10.0 mL + 25.0 mL = 35.0 mL

Now,

            C2 = (0.70 M x 10.0 mL) / 35.0 mL = 0.20 M

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