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The mass percent of Cl⁻ in a seawater sample is determined by titrating 25.00 mL of seawater with AgNO₃ solution, causing a precipitation reaction. An indicator is used to detect the end point, which occurs when free Ag⁺ ion is present in solution after all the Cl⁻ has reacted. If 63.30 mL of 0.2850 M AgNO₃ is required to reach the end point, what is the mass percent of Cl⁻ in the seawater (d of seawater = 1.024 g/mL)?

Respuesta :

Answer:

2.5 %

Explanation:

Considering:

[tex]Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}[/tex]

Or,

[tex]Moles =Molarity \times {Volume\ of\ the\ solution}[/tex]

Given :

For [tex]AgNO_3[/tex] :

Molarity = 0.2850 M

Volume = 63.30 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 63.30 × 10⁻³ L

Thus, moles of [tex]AgNO_3[/tex] :

[tex]Moles=0.2850 \times {63.30\times 10^{-3}}\ moles[/tex]

Moles of [tex]AgNO_3[/tex]  = 0.0180405 moles

Moles of [tex]AgNO_3[/tex]  = Moles of [tex]Cl^-[/tex]

Thus, Moles of [tex]Cl^-[/tex] = 0.0180405 moles

Molar mass of [tex]Cl^-[/tex] = 35.453 g/mol

Mass = Moles * Molar mass = 0.0180405 moles * 35.453 g/mol = 0.6396 g

Volume of sea water = 25.00 mL

Density = 1.024 g/mL

Density = Mass / Volume

Mass = Density * Volume = 1.024 g/mL * 25.00 mL = 25.6 g

[tex]Mass\ \%=\frac{Mass_{Chloride ion}}{Total\ mass}\times 100[/tex]

[tex]Mass\ \%=\frac{0.6396}{25.6}\times 100[/tex]

Mass percent of Cl⁻ = 2.5 %

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