A250 cm^3 solution containing 1,46 g of sodium chloride is added to an excess of silver nitrate solution. the reaction is given. nacl (aq)+agno, (aq)-agci(s)+nano, (aq) what is the concentration of the sodium chloride solution? (4) 7.1 calculate the mass of the precipitate. 7.2 (4) 18

The mass of the precipitate  that AgCl is 3.5803 g.

Explanation:

a) To calculate the molarity of solution, we use the equation:

We are given:

Mass of solute (NaCl) = 1.46 g

Molar mass of sulfuric acid = 58.5 g/mol

Volume of solution =

Putting values in above equation, we get:

0.09982 M is the concentration of the sodium chloride solution.

b)

Moles of NaCl =

according to reaction 1 mol of NaCl gives 1 mol of AgCl.

Then 0.02495 moles of NaCl will give:

of AgCl

Mass of 0.02495 moles of AgCl:

The mass of the precipitate  that AgCl is 3.5803 g.