How do you find the Initial Concentration with the information given? Within the
ID: 580027 • Letter: H
Question
How do you find the Initial Concentration with the information given? Within the experiment it states, "Note: In step 21, transfer the appropriate volume of sugar solutions (4.5 mL, 3.0 mL, and 6.0 mL). Also, use the diluted solution from the previous step as the initial concentration on the next dilution. Do not use the same initial concentration of 32% for all the dilutions.
Experiment 3 Exercise 3 Data Table 8 EE Data Table 9 Graph 1 Data Table 9: Solutions Density (g/mL) 1.115 g/ml Initial Concentration (% m/V) Volume Transferred (mL) 0 mL Final Concentration (% m/V) 32.0% Solution Volume (mL) Mass (g) 25 mL 27.88 g 25.16 g 24.96 g 24.79 24.81 32.0% 2.5 mL 4.5 mL 3.0 mL 25 mL 1.006 q/mL 3.2% 25 mL 998 g/mL 992 g/mL 992 g/mL 25 mL 25 mL 6.0 mLExplanation / Answer
Ans: To calculate the concentration of any solution, we can use a relative equation;
C1V1 = C2V2, where initial concentration of sample can be used in this equation to form the desired solution of known concentration or concentration can be calculated by the dilution of solution.
Example; for dilution of 0.2 M solution to 0.1 M
We can have 0.2×V1 = 0.1×V2
By considering the initial volume of 10 mL, we will have 0.2×10 = 0.1×V2
V2 = 20 mL
Hence 20 mL of solution need to be form from 10mL of 0.2M
Volume needed to form, b = V2-V1 = 10mL
We are keeping the total volume = 25 mL
Hence given data for solution 1, shows the initial solution concentration by m/V = 32%
Total volume = 25 mL, Volume transferred = 2.5 mL
Hence, 32% × 2.5mL = c × 25mL c = (32% × 2.5)/25 = 3.2 % final concentration (Given)
Hence for solution 2,
3.2% × 4.5mL = c × 25mL c = 0.5760 %
For solution 3,
0.576% × 3.0mL = c × 25mL c = 0.0691 %
For solution 4,
0.0691 % × 6.0mL = c × 25mL c = 0.0165 %
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