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STALIN [3.7K]
3 years ago
7

in Analytical chemistry laboratory. Nickle(ppm)in water sample estimated by two different students was as ,student 1 (25 24 24 2

4 24 28 26 ) and student 2 (24 25 24 24 25 25 ), if the true valude on Nickle in water was 20 ppm estimate the precision and accuracy of both students
Chemistry
1 answer:
frosja888 [35]3 years ago
6 0

Answer:

Answers are in the explanation

Explanation:

Precision and accuracy are both statistical concepts. Precision is about how close are the measured values to each other. Accuracy is closeness of the measured values to a specific value or reference value.

There are many ways to calculate precision and accuracy, one is stimating precision as <em>Relative Uncertainty </em>and accuracy as <em>Relative Error.</em>

Formulas are:

Relative Uncertainty: Standard desviation /  Measured value.

Relative Error: Measured Value - Expected Value / Expected value

<em>Measured value is the aveage of the values.</em>

<em />

For student 1:

Mean = 25

Standard desviation = 1.5275

Relative Uncertainty: 1.5275 /  25

Relative Uncertainty: 0.061

Relative Error: 25 - 20 / 20

Relative Error: 0.25

For student 2:

Mean = 24.5

Standard desviation = 0.5477

Relative Uncertainty: 0.5477 / 24.5

Relative Uncertainty: 0.02

Relative Error: 24.5 - 20 / 20

Relative Error: 0.225

<em>As the student 2 has a low Relative error and uncertainty, you can say student 2 is more accurate and precise</em>

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Explanation:

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In 1 L of solution 1.00 moles of calcium chloride is present.

Mass of solute or calcium chloride = m

m = 1 mol\times 111 g/mol = 111 g

Mass of solution = M

Volume of solution = V = 1L = 1000 mL

Density of solution , d= 1.07 g/mL

M=d\times V=1.07 g/mL\times 1000 mL=1,070 g

1) The value of %(m/M):

\frac{m}{M}\times 100=\frac{111 g}{1,070 g}\times 100=10.37\%

2) The value of %(m/V):

\frac{m}{V}\times 100=\frac{111 g}{1000 L}\times 100=11.1\%

Molality = \frac{\text{Moles of compound }}{\text{mass of solvent in kg}}

Normality=\frac{\text{Moles of compound }}{n\times \text{volume of solution in L}}

n = Equivalent mass

n = \frac{\text{molar mass of ion}}{\text{charge on an ion}}

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Moles of calcium ion = 1 mol (1 CaCl_2 mole has 1 mole of calcium ion)

n=\frac{40 g/mol}{2}=20

=\frac{1 mol}{20 g/mol\times 1L}=0.050 N

4) Normality of chlorine ions:

Moles of chlorine ion = 2 mol (1 CaCl_2 mole has 2 mole of chlorine ion)

n=\frac{35.5 g/mol}{1}=35.5

=\frac{2 mol}{35.5 g/mol\times 1L}=0.056 N

Moles of calcium chloride = 1.00 mol

Mass of solvent =  Mass of solution - mass of solute

= 1,070 g - 111 g = 959  g = 0.959 kg ( 1 g =0.001 kg)

5) Molality of the solution :

\frac{1 mol}{0.959 kg}=1.043 mol/kg

Moles of calcium chloride = n_1=1mol

Mass of solvent = 959 g

Moles of water = n_2=\frac{959 g}{18 g/mol}=53.28 mol

Mass of solvent = 959 g

6) Mole fraction of calcium chloride =

\chi_1=\frac{n_1}{n_1+n_2}=\frac{1mol}{1 mol+53.28 mol}=0.01842

7) Mole fraction of water =

\chi_2=\frac{n_2}{n_1+n_2}=\frac{53.28 mol}{1mol+53.28 mol}=0.9816

8) Mass of solution = m'

Volume of the solution= v = 100 mL

Density of solution = d = 1.07 g/mL

m'=d\times v=1.07 g/ml\times 100 g= 107 g

Mass of 100 mL of this solution 107 grams of solution.

9) Volume of solution = V = 100 mL

Mass of solution = M'' = 107 g

Mass of solute = m

The value of %(m/V) of solution = 11.1%

11.1\%=\frac{m}{100 mL}\times 100

m = 11.1 g

Mass of solvent = M''- m = 107 g -11.1 g = 95.9 g

95.9 grams of water was present in 100 mL of given solution.

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