1) A student forgets that the hexahydrate form of CoCl, is being used in this ex
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Question
1) A student forgets that the hexahydrate form of CoCl, is being used in this experiment and does the calculations using the molar mass of CoCla. How will that error show up in the unknown concentration? Why? 2) Your vernier instrument is very noisy- ie. it does not give a consistent measurement but fluctuates between values. What effect would that have on the standard deviation? ation from a calibration curve is y 4.397x. A sample of unknown concentration rbance of 2.313. What is the concentration of this substance? Based on the in the introduction, can we use this linear equation for this concentration? Explain.Explanation / Answer
1) the detail about the experiment is missed in this question. Eventhough, normally if the hexahydrate of CoCl2 is used, one has to use the molar mass of the hydrated salt. The molar mass of CoCl2 is only 129. 839 g/mol whereas that of CoCl2. 6H2O is 237. 930 g/mol which means there is mass difference by around 108 g/mol (i.e. mass of six water molecules). Most of the chemical experiment calculations are done using the mole concept. In this case number of mols measured by the student considering only CoCl2 will be higher than the number of mols when its hydrated form is considered. So the final result will have the error by weight %.
Number of mols = weight of (x)/atomic or molar mass of (x)
2) standard deviation Is the measure that summarises how far each measurement value is deviated from the mean or average.
For N measurements by a vernier instrument the standard deviation is measured as follows
1. Calculate the mean of N measurements
2. Find deviation of each value from the mean
3. Square these N deviations and add them
4. The sum is divided by N-1 and take the square root.
The standard deviation for any measurement can be large or small. What makes it larger or smaller is that how close is the dataset to the mean value. If the values are spread apart then its standard deviation is high. Here for the case of vernier instrument since it is too noisy and giving totally inconsistent values their standard deviation will be larger or increasing.
3)Absorbance (the measure of capacity of any substance to absorb the incident light) of any solution is directly propotional to its concentration. They are linearly related. Here the given caliberation curve has the equation y = 4.397 x where y must be the Absorbance and x must be the concentration.
Thus the concentration of unknown solution (x) = 2.313/4.397 = 0.526 M
The calculation is done assuming that the thickness or pathlength (l) of the solution and molar extinction coefficient (e) of the given solution is equal to unity.
Absorbance (A) = ecl + b ( Beer's law)
(since the concentration is independent variable which is taken as x -axis and Absorbance is dependent variable taken as y -axis and the graph which you get will be a straight line) and where, c is the concentration of the solution and b is the intercept.
Thus concentration of any unknown solution, C = (A-b)/el
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