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A certain solution contains 7 micro moles of Pb. What amount of solvent Pb should be diluted in order to attain a 3.09x10⁵ ppm of concentration? ​

Sagot :

Solution:

Step 1: Calculate the mass of Pb.

Note: 1 micromole = [tex]\it{1 × 10^{-6}}[/tex] mole

mass of Pb = moles of Pb × molar mass of Pb

mass of Pb = [tex]7 × 10^{-6}[/tex] mol × 207.2 g/mol

mass of Pb = 0.0014504 g

Step 2: Calculate the mass of solvent.

[tex]\text{parts per million} = \frac{\text{mass of Pb}}{\text{mass of Pb + mass of solvent}} × 10^{6}[/tex]

[tex]3.09 × 10^{5} = \frac{\text{0.0014504 g}}{\text{0.0014504 g + mass of solvent}} × 10^{6}[/tex]

[tex]0.309 = \frac{\text{0.0014504 g}}{\text{0.0014504 g + mass of solvent}}[/tex]

[tex]0.309(\text{0.0014504 g + mass of solvent}) = \text{0.0014504 g}[/tex]

[tex]\text{0.00044817 g + 0.309(mass of solvent) = 0.0014504 g}[/tex]

[tex]\text{0.309(mass of solvent) = 0.0014504 g} - \text{0.00044817 g}[/tex]

[tex]\text{0.309(mass of solvent) = 0.00100223 g}[/tex]

[tex]\text{mass of solvent} = \frac{\text{0.00100223 g}}{0.309}[/tex]

[tex]\text{mass of solvent = 0.00324 g}[/tex]

Note: 1 mg = [tex]\it{1 × 10^{-3}}[/tex] g.

[tex]\boxed{\text{mass of solvent = 3.24 mg}}[/tex]

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