Question

what is the mass of the precipitate formed when 50 ml of 16.9% solution of AgNO₃ is mixed with 50 ml of 5.8 % NaCl solution?


explain .
(class 11)

Answer 1

Well, w/v≡Mass of solute
Volume of solution, we should get over 7⋅g of silver chloride.

Explanation:

And thus mass of silver nitrate=16.9%×50⋅mL=8.45⋅g.

And this represents a molar quantity of 8.45⋅g169.87⋅g⋅mol−1

=0.0497⋅mol, with respect to AgNO3.

Likewise mass of sodium chloride=5.8%×50⋅mL=2.90⋅g.

And this represents a molar quantity of 2.90⋅g58.44⋅g⋅mol−1

=0.0497⋅mol with respect to NaCl.

Clearly the reagents are present in 1:1 molar ratio. The reaction that occurs in solution is the precipitation of a curdy white mass of AgCl(s), i.e. and we represent this reaction by the net ionic equation.....

Ag++Cl−→AgCl(s)⏐↓

Of course the complete reaction is....

AgNO3(aq)+NaCl(aq)→AgCl(s)⏐↓+NaNO3(aq), i.e. sodium nitrate remains in solution and can be separated (with effort) from the precipitate.

And given the stoichiometry, we gets 0.04974⋅mol×143.32⋅g⋅mol−1=7.11⋅g.



That's All,
Hope This Help,

Abhinav,
a Brainly User
:-)

Answer 2

Answer:

Let us calculate the moles of every compund.

Moles of AgNO3= 50∗16.9/100∗169.8

=0.05mole

Moles of NaCl = 50∗5.8/100∗58.5

=0.05mole

AgNo3 + NaCl ----> AgCl + NaNo3

Mass of AgCl = mole × molar mass

=0.05∗143.5

=7.16g

P.S.-Molar Mass of AgNO3 = 169.8 & NaCl = 58.5

HOPE THAT IT WILL HELP YOU MATE.