CHM 111/112 LAB MANUAL >C<
CHM 111/112 LAB MANUAL >C<
11th Edition
ISBN: 9781337310956
Author: SLOWINSKI
Publisher: CENGAGE LEARNING (CUSTOM)
Question
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Chapter 7, Problem 1ASA

(a)

Interpretation Introduction

Interpretation:

The mean value of the concentration for the given three values 0.02813 M, 0.02802 M, and 0.02788 M is to be calculated.

Concept introduction:

The mean value is defined as the average value of the number of observations. It is determined by doing the sum of the values of observations and then dividing it by overall observations.

(a)

Expert Solution
Check Mark

Answer to Problem 1ASA

The mean concentration is 0.02801 M.

Explanation of Solution

The mean value is calculated by doing the sum of the values of observations and then dividing it by the total number of observations. Mathematically, it can be calculated by the formula given below.

x¯=xin

Where,

  • x¯ is the mean value of the observations.
  • xi is the sum of observation values.
  • n is the number of observations taken.

The given values of the concentration are 0.02813 M, 0.02802 M, and 0.02788 M.

Substitute these values in the above formula to calculate the mean value of the concentration.

Meanconcentration=0.02813M+0.02802M+0.02788M3=0.084033=0.02801M

Therefore, the value of mean concentration is 0.02801M.

Conclusion

The mean value of the concentration is 0.02801 M.

(b)

Interpretation Introduction

Interpretation:

The standard deviation of the above result is to be calculated.

Concept introduction:

Standard deviation is defined as the numerical value which shows the deviation in the single observation value from the mean value of that observation data. The individual values of the observation and the mean value are used for the calculation of the standard deviation.

(b)

Expert Solution
Check Mark

Answer to Problem 1ASA

The standard deviation is 1.022×104M.

Explanation of Solution

The standard deviation determines the deviation of a single value from the mean value of the observations. The formula to calculate the standard deviation is given below.

s=(xi-x¯)2n

Where,

  • s is the value of standard deviation.
  • xi is the value of individual observation.
  • x ¯ is the value of mean concentration.
  • n is the value of the number of observations.

The mean concentration value is 0.02801 M. The individual concentration values are 0.02813 M, 0.02802 M, and 0.02788 M. The value of the number of observations is 3.

Substitute these values in the above formula.

s=(0.028130.02801)2+(0.028020.02801)2+(0.027880.02801)23=(0.00012)2+(0.00001)2+(0.00013)23=3.14×1083=1.022×104M

Conclusion

The standard deviation of the above result is 1.022×104M.

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Chapter 7 Solutions

CHM 111/112 LAB MANUAL >C<

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