Answer:
[Hg] = 3.4 x 10⁻⁶ M
Explanation:
Parts per million (ppm) is defined as (1 g solute) / (10⁶ g solution), or (10⁻⁶g solute) / (g solution). Here the density of the stream is 1.0 g/mL, so ppm is equal to 10⁻⁶g/mL.
The atomic mass of mercury is 200.59 g/mol, so the molarity is calculated as follows:
0.68ppm = (0.68 x 10⁻⁶g)/mL x (1000mL/L) x (mol/200.59g) = 3.4 x 10⁻⁶ M
Explanation:
It is given that there are 0.68 parts per million of mercury in the given stream.
This means that there are 0.68 mg of mass of mercury is present in [tex]10^{6}[/tex] mg of solution.
So, volume of solution will be calculated as follows.
Density = [tex]\frac{mass}{volume}[/tex]
1 g/ml = [tex]\frac{10^{-9} g}{volume}[/tex]
volume = [tex]1 \times 10^{-9} ml[/tex] (as 1 mg = 0.001 g)
As 1 ml = 0.001 L. So, [tex]1 \times 10^{-9} ml[/tex] will be equal to [tex]1 \times 10^{-12} L[/tex].
So, mass of mercury present in the stream = 0.68 ppm
= [tex]0.68 \times 10^{6}[/tex] g
Molar mass of mercury is 200 g/mol. As we know that,
No. of moles = [tex]\frac{mass}{\text{molar mass}}[/tex]
= [tex]\frac{0.68 \times 10^{-6}}{200 g/mol}[/tex]
= [tex]0.34 \times 10^{-8}[/tex] mol
Therefore, we will calculate the molarity of mercury as follows.
Molarity = [tex]\frac{\text{no. of moles}}{volume}[/tex]
= [tex]\frac{0.34 \times 10^{-8}}{1 \times 10^{-12} L}[/tex]
= [tex]0.34 \times 10^{4}[/tex]
Thus, molarity of mercury in the stream is [tex]0.34 \times 10^{4}[/tex].
