Answer:
0.052M
Step-by-step explanation:
We'll begin by writing the balanced equation for the reaction. This is given below:
AgNO3(aq) + KCl(aq) → AgCl(s) + KNO3 (aq)
Next, we shall determine the number of mole present in 0.684 g of precipitate, AgCl. This is illustrated below:
Mass of AgCl = 0.684g
Molar mass of AgCl = 108 + 35.5 = 143.5g/mol
Number of mole of AgCl =.?
Mole = Mass /Molar Mass
Number of mole of AgCl = 0.684/143.5
Number of mole of AgCl = 4.77×10¯³ mol
Therefore, 4.77×10¯³ mol of AgCl was produced.
Next, we shall determine the number of mole AgNO3 that reacted. This can be achieved by doing the following:
From the balanced equation above,
1 mole of AgNO3 reacted to produce 1 mole of AgCl.
Therefore, 4.77×10¯³ mol of AgNO3 will also react to produce 4.77×10¯³ mol of AgCl.
Next, we shall determine the molarity of AgNO3. This can obtain as follow:
Volume = 91.9mL = 91.9/1000 = 0.0919L
Mole of AgNO3 = 4.77×10¯³ mol
Molarity =.?
Molarity = mole /Volume
Molarity = 4.77×10¯³ / 0.0919
Molarity = 0.052M.
Therefore, the molarity of the AgNO3 is 0.052M.
Finally, we shall determine the molarity of silver ion in AgNO3. This can be obtained as follow:
AgNO3(aq) —> Ag+(aq) + NO3-(aq)
From the balanced equation above,
1 mole of AgNO3 produced 1 mole of Ag+.
Therefore, 0.052M AgNO3 will also produce 0.052M Ag+
Therefore, the molarity of the silver ion, Ag+ is 0.052M