Data and Calculations Table 1 Room Temperature = 24.7 C...

Data and Calculations Table 1 Room Temperature = 24.7 C Speed of sound = m/s Data Table 2 Data Table 3 Frequency Fork One = 512 Hz Frequency Fork Two = 1024 Hz L? (cm) L? (cm) L? (cm) L? (cm) L? (cm) L? (cm) L? (cm) L? (cm) 15.48 49.13 82.77 — 7.29 24.19 41.22 57.83 15.38 49.06 82.75 — 7.34 23.95 41.39 57.95 15.32 49.19 82.57 — 7.12 24.01 41.29 58.07 L? (m) L? (m) L? (m) L? (m) L? (m) L? (m) L? (m) L? (m) Calculations Table 2 L? = m L? = m L? = m L? = m x?? = m x?? = m x?? = m x?? = m i? =2(L?-L?) = m i? =(L?-L?) = m i? =2/3(L?-L?) = m x? = m x? = m V = f? = m/s % Err =

Transcript

00:01 Hello students, in this given problem, we have to find the amounts of tubcl remaining after each trials.

00:09 So four trials are given.

00:11 The solution is as the original solution of tbucl is 0 .1 molar, it has been diluted using the experiment by using 3 ml of it to make a reaction mixture of 10 ml.

00:26 So therefore, concentration of tbucl in the reaction mixture can be calculated as follows.

00:33 Molarity final into volume final is equal to molarity initial into volume initial.

00:40 Now substituting the given values, we have molarity final into 10 ml equals 0 .1 molar into 3 ml.

00:58 So molarity final equals 0 .1 molar into 3 ml divided by 10 ml.

01:14 Ml ml gets cancelled.

01:16 So we get 0 .03 molar.

01:20 So this is the concentration of tbcl in the reaction mixture.

01:26 Now, therefore, the starting concentration of tbucl in the reaction mixture is 0 .03 molar.

01:34 This is the same for each trial because all the trials are using 3 ml of original tbucl solution to make 10 ml of the reaction mixture.

01:45 Now for trial 1, as the reaction is completed to only 20%, this much of tbucl has been used during the reaction.

01:57 So therefore, percentage tbucl remaining is equal to 100 % minus 20%, which is equal to 80%.

02:17 As the volume of reaction does not change during the reaction, the remaining concentration of tbucl would be 80 % of the starting concentration.

02:27 So therefore, remaining tbucl concentration is equal to 80 % into 0 .03 molar, which is equal to 80 by 100 into 0 .03 molar.

02:54 So this is equal to 0 .024 molar.

03:00 So this is for trial 1.

03:03 Next for trial 2, as the reaction is completed to only 15%, it means this much of tbucl has been used during the reaction...

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