Answer:
a. ppb of trichloroethylene = 3 × 10⁶ ppb
b. ppm of Cl₂ = 3.8 ppm
c. Molarity = 0.0002 mol / L
d. Molarity = 0.0007 mol / L
e. For trace amount of concentrations
Step-by-step explanation:
a. Given data
mass of trichloroethylene = 25 mg
Volume of water = 9.5 L
ppb of trichloroethylene = ?
Solution
As we know that
1 L = 1000 milliliters
9.5 L = 9.5 × 1000
9.5 L = 9500 millileters (ml)
we consider 25 mg = 25 millileters
ppb = (mass of solute / mass of solvent) × 1000,000,000 (1 billion)
ppb of trichloroethylene = (25 ÷ 9500) × 1000,000,000
ppb of trichloroethylene = 0.003 × 1000,000,000
ppb of trichloroethylene = 3 × 10⁶ ppb
B. Given data
Mass of Cl₂ = 38 g
volume of water = 1.00 × 10⁴ L ( 10000 L)
ppm of Cl₂ = ?
Solution
Volume of water in ml = 1 L = 1000 ml
Volume of water in ml = 10000 × 1000
Volume of water in ml = 10000000 ml
we take 38 g = 38 ml
Now we convert it to ppm
ppm = (mass of solute / mass of solvent) × 1000000 (1 million)
ppm of Cl₂ = ( 38 ÷ 10000000 ) × 1000000
ppm of Cl₂ = 0.0000038 × 1000000
ppm of Cl₂ = 3.8 ppm
C. Given data
Concentration of F⁻ ( Fluoride ion) = 2.4 ppm
Molarity = ?
Solution
As we know that 1 ppm = 0.001 g / L
2.4 ppm = 2.4 × 0.001 g/L
2.4 ppm = 0.0024 g/L
Mass of flouride ions = 0.0024 g
Now we find number of moles
moles = mass / molar mass
molar mass of F⁻ = 19 g/mol
moles of F⁻ = 0.0024 g / 19 g/mol
moles of F⁻ = 0.0002 mol
Molarity = mol of solute / liter of solution
Molarity = 0.0002 mol / 1 L
Molarity = 0.0002 mol / L
D. Given data
Concentration of NO₃⁻ ( nitrate ion) = 45 ppm
Molarity = ?
Solution
As we know that 1 ppm = 0.001 g / L
45 ppm = 45 × 0.001 g/L
45 ppm = 0.045 g/L
Mass of nitrate ions = 0.045 g
Now we find number of moles
moles = mass / molar mass
molar mass of NO₃⁻ = 62 g/mol
moles of NO₃⁻ = 0.045 g / 62 g/mol
moles of F⁻ = 0.0007 mol
Molarity = mol of solute / liter of solution
Molarity = 0.0007 mol / 1 L
Molarity = 0.0007 mol / L
E. Reason of expressing concentration in ppm and ppb
Scientist prefer ppm and ppb notations when the concentration difference of solute and solvent are very high.
As water contains contaminants is a very low amount we can say in trace amounts so scientist prefer ppm and ppb rather than molarity.
Example
Arcenic is an under ground water contaminant and its concentration of 10 μg/L is dangerous for health.
Lets change this in to molarity
mass = 10 μg
10 μg = 10 / 1000000
10 μg = 0.00001 g
now find out moles of Arcenic
moles = mass / molar mass
molar mass of arcenic = 75 g/mol
moles = mass / molar mass
moles of arcenic = 0.00001 g / 75 g/mol
moles of arcenic = 0.00000012 mol
Molarity = moles of solute / litres of solution
Molarity = 0.00000012 mol / 1 L
Molarity = 0.00000012 mol/ L
As we can see that in molarity it is a negligible amount so scientists express it in ppm and ppb