Question

Water treatment plants commonly use chlorination to destroy bacteria. a byproduct is chloroform (chcl3), a suspected carcinogen produced when hocl, formed by the reaction of cl2 and water, reacts with dissolved organic matter. the united states, canada, and the world health organization have set a limit of 100. ppb of chcl3 in drinking water. convert this concentration into molarity, molality, mole fraction, and mass percent. enter your answers in scientific notation.

Answer 1

100. ppb of chcl3 in drinking water means  100 g of CHCl3 in 1,000,0000,000 g of water

Molarity, M

M = number of moles of solute / volume of solution in liters

number of moles of solute = mass of CHCl3 / molar mass of CHCl3

molar mass of CHCl3 = 119.37 g/mol

number of moles of solute = 100 g / 119.37 g/mol = 0.838 mol

using density of water = 1 g/ ml => 1,000,000,000 g = 1,000,000 liters

M = 0.838 / 1,000,000 = 8.38 * 10^ - 7 M <----- answer

Molality, m

m = number of moles of solute / kg of solvent

number of moles of solute = 0.838

kg of solvent = kg of water = 1,000,000 kg

m = 0.838 moles / 1,000,000 kg = 8.38 * 10^ - 7 m <----- answer

mole fraction of solute, X solute

X solute = number of moles of solute / number of moles of solution

number of moles of solute = 0.838

number of moles of solution = number of moles of solute + number of moles of solvent

number of moles of solvent = mass of water / molar mass of water = 1,000,000,000 g / 18.01528 g/mol = 55,508,435 moles

number of moles of solution = 0.838 moles + 55,508,435 moles = 55,508,436 moles

X solute = 0.838 / 55,508,435 = 1.51 * 10 ^ - 8 <------ answer

mass percent, %

% = (mass of solute / mass of solution) * 100 = (100g / 1,000,000,100 g) * 100 =

% = 10 ^ - 6 % <------- answer

Answer 2

Molarity :M =  8.368.10⁻⁷

molality = 8.368.10⁻⁷

mole fraction  = 1.508 .10⁻⁸

mass percent = 10⁻⁵%

Further explanation

The concentration of a substance can be expressed in several quantities such as moles, percent (%) weight/volume,), molarity, molality, parts per million (ppm) or mole fraction. Concentration shows the amount of solute in a unit of the amount of solvent.

• Molality (m)

Molality shows how many moles are dissolved in every 1000 grams of solvent.

$m=\frac{n}{p}$  

m = Molality

n = number of moles of solute

p = solvent mass (1000 grams)

• Mole

Mole itself is the number of particles contained in a substance

amounting to 6.02.10 ^ 23

Mole can also be sought if the amount of substance mass and its molar mass is known

$mole=\frac{mass}{molar mass}$

• Molarity (M)

Molarity is a way to express the density of the solution

Molarity shows the number of moles of solute in every 1 liter of solute or mmol in each ml of solution

$\large {\boxed {\bold {M ~ = ~ \frac {n} {V}}}$

Where

M = Molarity

n = Number of moles of solute

V = Volume of solution

ppb mean : part per billion or 1 : 10⁹, the convertion are :

• 1. molarity

1 ppm = 1 mg/l

1 ppm = 1000 ppb

1 ppb = 10⁻³ mg/l

100 ppb = 0.1 mg/l = 10⁻⁴ g/l

molar mass CHCl₃ = 12 + 1 + 35.5 .3 = 119.5 grams/mole

mole CHCl₃ = mass : molar mass

mole CHCl₃ = 10⁻⁴ g : 119.5

mole CHCl₃ = 8.368.10⁻⁷

Because the water(solvent) = 1 L, so the molarity :

M = mol : L

M = 8.368.10⁻⁷ : 1 L

M =  8.368.10⁻⁷

• 2. molality

mole CHCl₃ = 8.368.10⁻⁷

Solvent = 1 L = 1 Kg(ρ water = 1 g/ml or 1 kg/L)

molality = mole : 1 kg solvent

molality = 8.368.10⁻⁷  : 1

molality = 8.368.10⁻⁷

• 3. mole fraction

mole water = 1000 gr : 18 (molar mass H₂O)

mole water = 55.5

mole fraction = mole CHCl₃ : (mole H₂O + mole CHCl₃)

mole CHCl₃ is considered very small, then

mole fraction = mole CHCl₃ : mole H₂O

mole fraction = 8.368.10⁻⁷ : 55.5

mole fraction  = 1.508 .10⁻⁸

• 4. mass percent

mass CHCl₃ = 10⁻⁴ grams

mass H₂O = 1000 grams

mass percent = mass CHCl₃ : (mass CHCl₃+mass H₂O) x 100%

mass CHCl₃ is considered very small, then

mass percent = (mass CHCl₃ : mass H₂O)x 100%

mass percent = 10⁻⁴  : 10³ x 100%

mass percent = 10⁻⁵%

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Keywords: molality, molarity, mass percent, mole fraction