To 40cm^3 of 1mol AgNO3 is added 20cm^3 of 0.500 AlCl3. What is the molar concentration of the resulting AgNO3 solution.
The reaction, if needed, is
3AgNO3 + AlCl3 ==> 3AgCl + Al(NO3)3
I worked out the number of moles present in AgNO3 which is 0.04 mol, and the number of moles of AlCl3 which is 0.01 mol.
What do I do next though?
Thanks in advance! Any help is greatly appreciated
Question: A solution of Silver nitrate, AgNO3, contains 1.08 grams in 250 cm3. What is its molarity?
Solution:
Molar mass of Silver nitrate = 108 g/mole.
n(AgNO3) = 1.08 grams of AgNO3 x 1 mole of AgNO3108 grams AgNO3 = 0.01 mole.
Number of moles in 1 dm3 = 0.01 mole x 1000/250 = 0.04 moles/dm3
So, Molarity of solution = 0.04M.
This is an example, if 0.04 mol is right then replace 250 with 40 and that's your answer.
To find the molar concentration of the resulting AgNO3 solution, you need to determine the total volume of the solution and then divide the number of moles of AgNO3 by that volume.
Here's how you can proceed:
1. Determine the total volume of the resulting solution. You have initially mixed 40 cm^3 of AgNO3 and 20 cm^3 of AlCl3, so the total volume is 40 cm^3 + 20 cm^3 = 60 cm^3.
2. Calculate the number of moles of AgNO3 (nAgNO3). You already calculated this to be 0.04 mol.
3. Divide the number of moles of AgNO3 by the total volume of the solution.
Molar concentration (C) = nAgNO3 / V
where nAgNO3 is the number of moles of AgNO3 and V is the volume of the solution. In this case, since the volume is given in cm^3, your final concentration will be in moles per cubic centimeter (mol/cm^3) or molarity (M).
C = 0.04 mol / 60 cm^3 = 0.00067 mol/cm^3 = 0.00067 M
Therefore, the molar concentration of the resulting AgNO3 solution is 0.00067 M.
Note: It's important to ensure that your units are consistent throughout the calculation. In this case, the units for volume and concentration must be cubic centimeters (cm^3) to match the given information.