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A chemistry student is given a 25 mL sample of a solution containing an unknown amount of sodium chloride. After adding excess silver nitrate solution to the sample, the student records the mass of the resulting precipitate to be 2.655 grams. Using mass balance calculations, determine the mass percentage of sodium chloride in the original solution.

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To determine the mass percentage of sodium chloride  NaCl  in the original solution, we first need to find the moles of the precipitate formed, which is silver chloride  AgCl .The balanced chemical equation for the reaction between sodium chloride and silver nitrate is:NaCl aq  + AgNO aq   AgCl s  + NaNO aq From the equation, we can see that the mole ratio between NaCl and AgCl is 1:1.First, we need to find the moles of AgCl formed. The molar mass of AgCl is  107.87 g/mol for Ag  +  35.45 g/mol for Cl  = 143.32 g/mol.Moles of AgCl = mass of AgCl / molar mass of AgClMoles of AgCl = 2.655 g / 143.32 g/mol = 0.0185 molSince the mole ratio between NaCl and AgCl is 1:1, the moles of NaCl in the original solution is also 0.0185 mol.Now, we can find the mass of NaCl in the original solution. The molar mass of NaCl is  22.99 g/mol for Na  +  35.45 g/mol for Cl  = 58.44 g/mol.Mass of NaCl = moles of NaCl  molar mass of NaClMass of NaCl = 0.0185 mol  58.44 g/mol = 1.080 gNow, we can find the mass percentage of NaCl in the original solution. The total mass of the solution is not given, so we will assume that the density of the solution is approximately equal to the density of water  1 g/mL  since the concentration of NaCl is not very high. Therefore, the mass of the 25 mL solution is approximately 25 g.Mass percentage of NaCl =  mass of NaCl / total mass of solution   100Mass percentage of NaCl =  1.080 g / 25 g   100 = 4.32%The mass percentage of sodium chloride in the original solution is approximately 4.32%.

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