170 grams of silver nitrate solution was mixed with an excess of sodium chloride solution

170 grams of silver nitrate solution was mixed with an excess of sodium chloride solution, and a precipitate weighing 8.61 grams was formed. calculate the mass fraction of salt in the silver nitrate solution.

Given:

m (AgNO3) solution = 170 g

m (sediment) = 8.61 g

To find:

w (AgNO3) =?

Decision:

Let’s compose the reaction equation:

AgNO3 + NaCl = AgCl (precipitate) + NaNO3.

Let’s calculate the amount of substance for AgCl:

n (AgCl) = m (AgCl) / M (AgCl) = 8.61 g / (108 + 35.5) g / mol = 0.06 mol.

It can be seen from the reaction equation that:

n (AgCl) = n (AgNO3).

Knowing the amount of AgNO3 substance, we find the mass of AgNO3 salt in solution:

m (AgNO3) = n (AgNO3) * Mr (AgNO3) = 0.06 mol * (108 + 14 + 16 * 3) g / mol = 10.2 g.

We find the mass fraction of salt in the solution by the formula:

w (AgNO3) = m (AgNO3) / m (AgNO3) p-p * 100% = 10.2 g / 170 g * 100% = 6%.

Answer: 6%.