Question

What is the mass of precipitate formed when a 50mL of 16.9% w/v solution of $AgN{O_3}$ is mixed with 50mL of 5.8% w/v $NaCl$ solution?

Answer 1

The term percent w/v is known weight per volume, it is the amount of solute (in grams) dissolved in 100 millilitres of solution. This term provides information about the concentration of the solution.

Complete answer:
We are given two solutions, one is 50mL of 16.9% w/v solution of $AgN{O_3}$ and the second is 50mL of 5.8% w/v $NaCl$ solution. We’ll first find out the amount of solute (in grams) dissolved in each solution.
$AgN{O_3}$ solution: Volume taken = 50mL and %w/v = 16.9%
Amount of $AgN{O_3}$ present $= \dfrac{{16.9}}{{100}} \times 50 = 8.45g$
No. of moles of $AgN{O_3}$ will be $= \dfrac{{Mass}}{{M.{M_{AgN{O_3}}}}} = \dfrac{{8.45g}}{{169.87g/mol}} = 0.0497mol$ of $AgN{O_3}$ in the solution
(Molar mass of $AgN{O_3}$ is 169.87 g/mol)
$NaCl$ solution: Volume taken = 50mL and %w/v = 5.8%
Amount of $NaCl$ present $= \dfrac{{5.8}}{{100}} \times 50 = 2.90g$
No. of moles of $NaCl$ will be $= \dfrac{{Mass}}{{M.{M_{NaCl}}}} = \dfrac{{2.90g}}{{58.44g/mol}} = 0.0497mol$ of $NaCl$ in the solution
(Molar mass of $NaCl$ is 58.44 g/mol)
The no. of moles of $AgN{O_3}$ = No. of moles of $NaCl$ , hence the molar ratio is $1:1$. The complete reaction can be given as: $AgN{O_3} + NaCl \to AgCl \downarrow + NaN{O_3}$
AgCl is precipitated and it’s no. of moles will also be equal to that of $AgN{O_3}$ and $NaCl$.
The molar mass of AgCl is 143.32g/mol and no. of moles of AgCl is 0.0497moles
Therefore, the amount of AgCl (in grams) can be calculated as: $Moles \times M.{M_{AgCl}} = 0.0497 \times 143.32 = 7.11g$
Therefore, the mass of precipitate formed will be 7.11 grams.

Note:
Silver halides are very tough to separate out from a mixture. The particle size of these halides are very small. They can be separated by filtration, but it has been carried out very slowly. It likely clogs the pores of the filter. AgCl has a photoactive nature, i.e., it decomposes when exposed to light.