Question

A solution contains an unknown mass of dissolved silver ions. When potassium chloride is added to the solution, a white precipitate forms. The precipitate is filtered and dried and then found to have a mass of 245 mg. What mass of silver was in the original solution? (Assume that all of the silver was precipitated out of solution by the reaction.)

Answer 1

Answer:

Mass of silver in the original solution = 0.1845 g

Explanation:

The precipitate must be of silver chloride. So,

Given that:

Mass of silver chloride = 245 mg = 0.245 g ( 1 mg = 0.001 g )

Molar mass of silver chloride = 143.32 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$Moles= \frac{0.245\ g}{143.32\ g/mol}$

$Moles= 0.00171\ mol$

From the reaction,  

$KCl+Ag^+\rightarrow AgCl+K^+$

1 mole of AgCl is formed from 1 mole of silver

So,

0.00171 mole of AgCl is formed from 0.00171 mole of silver

Moles of silver = 0.00171 moles

Molar mass of silver = 107.8682 g/mol

Mass = Moles*Molar mass = 0.00171 moles*107.8682 g/mol = 0.1845 g

Mass of silver in the original solution = 0.1845 g