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 %

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the ideal gas law equation is as follows

pv = nrt

where p - pressure

v - volume

n - number of moles

r - universal gas constant

t - temperature

so if the pressure, volume and temperature are already known

we are left with n and r

since r is the universal gas constant that has a known fixed value then r too is known

so we are left with 'n'

once we know temperature volume and pressure

we can find the number of moles of gas present using the ideal gas law equation

this is not my own answer, this is an answer for the same question by the user blahblahmali. i do not claim ownership of this account.