Question

The concentration of the alkane c20h42 (fm 282.56) in a particular sample of rainwater is 0.2 ppb. assume that the density of rainwater is close to 1.00 g/ml and find the molar concentration of c20h42.

Answer 1

The molar concentration of C20H42 in a sample of rainwater with a concentration of 0.2 ppb and a density of 1.00 g/mL is calculated to be 7.08x10^-10 M.

Step-by-step explanation:

To find the molar concentration of C20H42 in a sample of rainwater with a concentration of 0.2 ppb (parts per billion), we start by understanding that 0.2 ppb means there are 0.2 grams of C20H42 for every billion grams of rainwater. Since the density of rainwater is approximately 1.00 g/mL, 1 liter (1000 mL) of rainwater would have a mass of 1000 grams. We can then convert the concentration from ppb to grams per liter (g/L).

0.2 ppb is equivalent to 0.2 grams of C20H42 per 1,000,000,000 grams of water, which, in terms of liters of water (assuming 1 g/mL density), would be 0.2 grams C20H42 per 1,000 liters of rainwater. Therefore, to find the concentration in g/L, we perform the following calculation:

0.2 g C20H42 / 1,000,000 g water * 1,000 g water / L = 2x10^-7 g/L

To convert this to molar concentration, we divide the mass by the molar mass of C20H42:

(2x10^-7 g/L) / (282.56 g/mol) = 7.08x10^-10 mol/L

The molar concentration of C20H42 in rainwater is thus 7.08x10^-10 M.

Answer 2

Formula used for determining the molarity is:

`Molarity = (number of moles of solute)/(volume of solution in Liters))` -(1)

Number of moles of solute that is
`C_(20)H_(42)` is determined as:

`number of moles = (given weight)/(Molar mass)` -(2)

Mass of the solute can be calculated as:

`Density = (mass)/(volume)`

`mass = density* volume` - (3)

density of rain water = 1.00 g/mL (given)

Concentration of
`C_(20)H_(42)` = 0.2 ppb (given)

Since,
`1 ppb =`
`10^(-3) mg/L`

So, 0.2 ppb =
`0.2* 10^(-3) mg/L = 2* 10^(-4) mg/L`

Substituting the values in formula (3):

Mass =
`2* 10^(-4) mg = 2* 10^(-7) g` (As 1 mg =
`10^(-3)` g).

Now, substituting the values of mass and formula mass in formula (2):

Number of moles =
`(2* 10^(-7))/(282.56) = 7.078* 10^(-10) mole`

Volume of solution = 1 L

Substituting the values of number of moles and volume of solution in formula (1):

Molarity =
`(7.078* 10^(-10) mole)/(1 L) = 7.078* 10^(-10) M`