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)?

2.5 %

Explanation:

Considering:

Or,

Given :

For :

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 :

Moles of  = 0.0180405 moles

Moles of  = Moles of

Thus, Moles of = 0.0180405 moles

Molar mass of = 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 percent of Cl⁻ = 2.5 %

c) should be the answer.