The mass percentage of chloride ion in a 25 ml sample of seawater was deteremined by titrating the saample with silver nitrate , precipitating silver chloride. it took 42.58 ml of 0.2997 m silver nitrate soltuion to reach the equivalence point in the titration. what is the mass percentage of chloride ion in seawater if its density is 1.025g/ml

The balanced chemical equation is expressed as:

Cl-(aq) + AgNO3(aq) ---> AgCl(s) + NO3-(aq) 

To determine the mass percent of chloride ion in the sample, we need to calculate for the mass of chloride ion present in the sample. We start from the volume and concentration of AgNO3 that was used to titrate. 42.58 mL AgNO3 (1L / 1000mL) (0.2997 M) (1 mol Cl- / 1 mol AgNO3) (35.45 g Cl- / 1 mol Cl-) = 0.4524 g Cl-
Then, we calculate for the total mass of the sample from the density and the volume,
Mass of sample = 1.025 g / mL ( 25.0 mL ) = 25.625 grams sample
Percent Cl- = 0.4524 g Cl- / 25.625 g sample x 100 = 1.77% Cl-

The answer is frictional force

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