how do you add molarity?

Question

so i have solution A:

3.00L of 3.00M of NaCl

and solution B:

2.00L of 2.00 M of AgNO3

if i mix these two solutions together, how do i calculate the concentrations (in M) of the following ions in solutions C.

Cl minus?

Answer 1

You have to re-calculate the molar concentrations. So, for the sodium chloride, you are effectively diluting the initial amount of NaCl into a larger volume. So, (3L X 3.0mol/L)/5.0 L will give you the final concentration of sodium and chloride ions. You can repeat this calculation with the silver nitrate solution.

When you mix these solutions, though, silver chloride will precipitate. The final concentration of silver ions will be determined by the remaining concentration of chloride ions and by the solubility product of silver chloride.

I'll be happy to help with this, if you'd like...Send me a message.

Answer 2

First, you have to calculate now many moles of Cl- are there in solution A

3L of 3 M Cl-

M=moles/liters
so moles = liters*M or 3*3=9 moles

now, you have to calculate the total volume of water
3.00L+2.00L=5.00L

then, to calculate the concentratino again:
M=moles/liters
M=9/5
M=1.8

Answer 3

molar mass of AlBr3= 27 + (seventy 9.9*3)=266.7g/mol 300mL to litres=3 hundred/one thousand= 0.3Litres mole= concentration * quantity in litres =0.158M* 0.3 =0.0474mol of AlBr3 mass =mol * molar mass =0.0474mol*266.7g/mol =12.6 grams of AlBr3 sorry, do not recognize any web content,yet wish u understand what i did.

Answer 4

NaCl + AgNO3--------NaNO3 + AgCl
M * V = # moles
initial
9.00 moles + 4.00 moles--------0 + 0
final
5.00 moles + 0 --------- 4.00 moles + 4.00 moles

5moles of Cl- from excess NaCl + 4moles of Cl- from the product AgCl
9 moles in total
total volume is 3.00+2.00=5.00L
M=#moles/volume
9.00/5.00=1.8 M