If your solution’s final volume of 32.55 mL yields 3.6 grams of AgCl, determine the Molarity of the final solution. Also, how many milliliters of a 1.85 M solution of AgNO3 did you add to an excess of
aqueous NaCl to provide the 3.6 grams of AgCl?
I don't understand the question.
Actually this question is two parts.
The first part is to find the Final Molarity which is
(3.6 g * 1 mole /143.35 g) / (32.55 ml * 1 L /1000 ml) = 0.77 mol/L or M
The second part is to find how many millilitres of AgNo3 which has 1.85 M.
I don't understand how to do this second part.
Please help me this.
To determine the molarity of the final solution, we need to use the formula:
Molarity (M) = moles of solute / volume of solution in liters
First, let's calculate the moles of AgCl using the given mass of 3.6 grams and the molar mass of AgCl, which is 143.32 g/mol:
moles of AgCl = mass of AgCl / molar mass of AgCl
moles of AgCl = 3.6 g / 143.32 g/mol ≈ 0.0251 mol
Next, we need to convert the final volume from milliliters to liters:
volume of solution = 32.55 mL = 32.55 mL / 1000 mL/L = 0.03255 L
Now we can calculate the molarity of the final solution:
Molarity = moles of AgCl / volume of solution
Molarity = 0.0251 mol / 0.03255 L ≈ 0.771 M
Therefore, the Molarity of the final solution is approximately 0.771 M.
Now let's determine the volume of the 1.85 M solution of AgNO3 that was added to the excess of aqueous NaCl to provide 3.6 grams of AgCl.
From the balanced chemical equation for the reaction between AgNO3 and NaCl, we know that 1 mole of AgNO3 reacts with 1 mole of NaCl to form 1 mole of AgCl:
AgNO3 + NaCl -> AgCl + NaNO3
Since 1 mole of AgNO3 reacts to form 1 mole of AgCl, the moles of AgNO3 used can be calculated as follows:
moles of AgNO3 = moles of AgCl = 0.0251 mol
Now we can calculate the volume of the 1.85 M solution of AgNO3 using the formula:
Molarity (M) = moles of solute / volume of solution in liters
volume of AgNO3 = moles of AgNO3 / Molarity of AgNO3
volume of AgNO3 = 0.0251 mol / 1.85 mol/L ≈ 0.0135 L
Finally, let's convert the volume from liters to milliliters:
volume of AgNO3 = 0.0135 L * 1000 mL/L ≈ 13.5 mL
Therefore, approximately 13.5 milliliters of the 1.85 M solution of AgNO3 were added to an excess of aqueous NaCl to provide the 3.6 grams of AgCl.