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

Formula used for determining the molarity is:

$Molarity = \frac{number of moles of solute}{volume of solution in Liters)}$  -(1)

Number of moles of solute that is $C_{20}H_{42}$ is determined as:

$number of moles = \frac{given weight}{Molar mass}$  -(2)

Mass of the solute can be calculated as:

$Density = \frac{mass}{volume}$

$mass = density\times volume$ - (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 = $0.2\times 10^{-3} mg/L = 2\times 10^{-4} mg/L$

Substituting the values in formula (3):

Mass = $2\times 10^{-4} mg = 2\times 10^{-7} g$      (As 1 mg = $10^{-3}$ g).

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

Number of moles = $\frac{2\times 10^{-7}}{282.56} = 7.078\times 10^{-10} mole$

Volume of solution = 1 L

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

Molarity = $\frac{7.078\times 10^{-10} mole}{1 L} = 7.078\times 10^{-10} M$