Question

Three student have made multiple measurements of solubility of sodium chloride (in g of NaCl per 100g of water) . their result are summarized in the table below the correct value for the solubility of sodium chloride is 35.99. which of the following student had a better precision and which one had a better accuracy than all show your working

Answer 1

To determine which student had better precision and accuracy, we need to compare their measurements to the correct value of sodium chloride's solubility (35.99 g NaCl/100g water).

Precision refers to how close the measurements are to each other. To determine which student had better precision, we can calculate the average deviation for each student's measurements.

Let's assume the measurements made by the three students are as follows:

Student 1: 34.50, 34.75, 34.80
Student 2: 35.80, 36.20, 35.90
Student 3: 35.95, 35.98, 36.02

We can calculate the average deviation for each student by taking the absolute value of the difference between each measurement and the average of all measurements, and then calculating the average of these absolute differences.

For Student 1:
Average deviation = (|34.50 - 34.68| + |34.75 - 34.68| + |34.80 - 34.68|) / 3
= (0.18 + 0.07 + 0.12) / 3
= 0.13 / 3
≈ 0.043 g NaCl/100g water

For Student 2:
Average deviation = (|35.80 - 35.97| + |36.20 - 35.97| + |35.90 - 35.97|) / 3
= (0.17 + 0.23 + 0.07) / 3
= 0.47 / 3
≈ 0.157 g NaCl/100g water

For Student 3:
Average deviation = (|35.95 - 35.98| + |35.98 - 35.98| + |36.02 - 35.98|) / 3
= (0.03 + 0 + 0.04) / 3
= 0.07 / 3
≈ 0.023 g NaCl/100g water

Now, let's determine accuracy by comparing each student's average measurement to the correct value.

For Student 1:
Average measurement = (34.50 + 34.75 + 34.80) / 3
= 34.68 g NaCl/100g water
Accuracy = |34.68 - 35.99|
≈ 1.31 g NaCl/100g water

For Student 2:
Average measurement = (35.80 + 36.20 + 35.90) / 3
= 35.97 g NaCl/100g water
Accuracy = |35.97 - 35.99|
≈ 0.02 g NaCl/100g water

For Student 3:
Average measurement = (35.95 + 35.98 + 36.02) / 3
= 35.98 g NaCl/100g water
Accuracy = |35.98 - 35.99|
≈ 0.01 g NaCl/100g water

Based on the calculations, Student 3 had better precision as they had the lowest average deviation (0.023 g NaCl/100g water) compared to Student 1 (0.043 g NaCl/100g water) and Student 2 (0.157 g NaCl/100g water).

Additionally, Student 3 had better accuracy as their average measurement (35.98 g NaCl/100g water) was closest to the correct value (35.99 g NaCl/100g water), with an accuracy of 0.01 g NaCl/100g water. Student 2 also had good accuracy with an accuracy of 0.02 g NaCl/100g water. However, Student 1 had lower accuracy with an accuracy of 1.31 g NaCl/100g water.