Environmental Measurements ReviewName

Atomic mass for different elements: C – 12, O – 16, H – 1, Na – 23, Cl – 35.5, As – 74.9

Molar Mass = The molar mass of a compound is equal to the sum of the atomic masses of its constituent atoms in g/mol.

Mole: mole = mass ÷ molar mass mass = moles × molar mass

Mole = The amount of substance that contains as many elementary entities as there are atoms in 12 g of carbon-12.One mole (abbreviated mol) is equal to 6.022×1023 molecular entities (Avogadro's number), and each element has a different molar mass depending on the weight of 6.022×1023 of its atoms (1 mole).

Molarity, M =moles solute/liter of solution

Weight%, Wt% =mass ratio × 100% =mass of solute/mass of solution × 100%

Parts per million, ppm = mass ratio× 106 = mass of solute/mass of solution × 106(The ppm concept is a similar to that of percent, which is essentially parts per hundred)

In aqueous solution, 1 ppm = 1mg/L

In air, under STP condition (1 atm, 25oC) c (mg/m3) = ppm × MW / 24.5 (MW: molar mass or molecular weight)

In air, under nonstandard condition c (µg/m3) = ppm × MW × P × 1000 / (RT) (R: 0.08206 liter-atm/mole-K; T is the temperature in K)

  1. 1 ppm = 0.0001 %
  1. How many moles of Ca(NO3)2 are there in 75 mL of 0.25 M solution?

0.25 mol/L × 0.075L = .01875mol

  1. What mass of NaCl are dissolved in 152 mL of a solution if the concentration of the solution is 0.364 M?

0.364 × (23 + 35.5) × 0.152 = 3.24 g

  1. Carbon monoxide (CO) affects the oxygen-carrying capacity of your lungs. Exposure to 50 ppmv CO for 90 min has been found to impair one's ability to discriminate stopping distance; therefore, motorists in heavily polluted areas may be more prone to accidents. Are motorists at a greater risk of accidents if the CO concentration is 55 mg/ m3? Assume a temperature of 298 K and 1 atm.

No

55 mg/ m3 = ppm × 28 (MW of CO) / 24.5ppmv = 48 50ppmv

  1. A dry cleaning facility owned by JMA Inc. has been observed to have impacted 6,000 gallons of groundwater with 0.70 lb. of tetrachloroethylene (PCE). Assuming all the PCE is present in the dissolved phase and the chemical pollutant is evenly distributed throughout the impacted volume of the groundwater, what is the concentration of PCE in groundwater in ppm?

{0.7 lb. × (1000g/2.2 lb.) / [6000 gal. × (3.785 L/1 gal.)]} × (1000 mg / 1 g) = 0.014 mg/L = 14 ppm

  1. Convert 800 mg/m3 O3 (w/v) to ppm (v/v) at 20oC.

[Hint: Use the equation connecting mg/m3 to ppm]

No. moles O3 in 1 m3 air = 800 mg / 48 g mol-1 =800 × 10-3 g / 48 g mol-1

= 1.67 × 10-2 mol

Volume occupied by 1 mole at 20 °C and 1 atm (SATP)

= RT = 0.08206 × 293 = 24.0 L = 0.0240 m3

Volume O3 in 1 m3 air

= 1.67 × 10-2 mol × 0.0240 m3 mol-1

= 400 × 10-6 m3 = 400 ppmv

Alternative solution: 800 mg/m3 = 800 × 103µg/m3 = ppmv × MW × 1000 / (0.08206× 293)

  1. Which of the following is NOTa true statement?
  1. ppm = µg / g in solid
  2. ppm = mg /g in solid
  3. ppm = mg / L in water
  4. ppm = µg / mL in water
  1. Which of the following is a true statement about concentration in the gas phase?
  1. ppmv = 1 mg / m3
  2. ppmv = 1 L / 1 m3
  3. ppmv = 1 cm3 / 1 L
  4. ppmv = 1 mL / 1 m3
  1. A solution is made by mixing 250.0 g of hexane and 50.0 g of octanol. What is the mass percent of the octanol?
  1. 83.3%
  2. 16.7%
  3. 20%
  4. None of the above

50 / (250 + 50) = 1/6 = 16.7%

  1. What is the molarity of a solution prepared by diluting 43.72 mL of 5.005 M aqueous K2Cr2O7to 500 mL?
  1. 0.0044 M
  2. 0.0879 M
  3. 0.438 M
  4. 0.870 M.

43.72 mL× 10-3× 5.005 M/ .500 L = 0.438

  1. What volume of a concentrated solution of potassium hydroxide (6.00 M) must be diluted to 200. mL to make a 0.880 M solution of potassium hydroxide?
  1. 26.4 mL
  2. 29.3 mL
  3. 50.0 mL
  4. 176 mL

200 mL × 0.88 = V × 6

  1. The Henry's law constant, H, is used to describe a chemical's equilibrium between the air and water (often termed the dissolved or aqueous) phases. Calculate the concentration of dissolved oxygen (DO) (units of moles/L and mg/L) in a water equilibrated with the atmosphere at 25°C. The Henry's law constant for oxygen at 25oC is 1.29 × 10-3 mole/L-atm.[Hint: The partial pressure of oxygen in the atmosphere is 0.21 atm.]

1.29 × 10-3 mole/L-atm × 0.21 atm = 2.71 × 10-4 mole/L

2.71 × 10-4 mole/L×32g/mole = 8.7 mg/L

  1. A water sample has been found to contain 0.6 mM of As. The drinking water guidelines for arsenicis 10 ppb. Does thissample exceed the drinking water guidelines for arsenic?

Yes

0.6 mM = 0.6 × 10-3mol/L × 74.9 g/mol × 1 L/1000g = 0.00004494 = 44.94 ppm = 44940 ppb

  1. The total volume of the oceans on Earth is 1.35 × 1018 m3. What are the masses of the following elements in the Ocean in units of kg?

Oxygen, O, for which the concentration in the seawater is 857,000 ppm

1.35 × 1018 m3 = 1.35 × 1021 L

857,000 ppm = 857,000 mg/L

Mass of oxygen = 857,000 mg/L × 1.35 × 1021 L = 1156950 × 1021 mg = 1.16 × 1021 kg