Question

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)?

Answer 1

Answer:

2.5 %

Explanation:

Considering:

$Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}$

Or,

$Moles =Molarity \times {Volume\ of\ the\ solution}$

Given :

For $AgNO_3$ :

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 $AgNO_3$ :

$Moles=0.2850 \times {63.30\times 10^{-3}}\ moles$

Moles of $AgNO_3$  = 0.0180405 moles

Moles of $AgNO_3$  = Moles of $Cl^-$

Thus, Moles of $Cl^-$ = 0.0180405 moles

Molar mass of $Cl^-$ = 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

$Mass\ \%=\frac{Mass_{Chloride ion}}{Total\ mass}\times 100$

$Mass\ \%=\frac{0.6396}{25.6}\times 100$

Mass percent of Cl⁻ = 2.5 %