0.0981 mol NaCl x (1 mol Cl / 1 mol NaCl) = 0.0981 mol Cl
0.00512 mol MgCl2 x (2 mol Cl / 1 mol MgCl2) = 0.01024 mol Cl
Step 4. Add the amount of moles of Cl and divide by volume of solution in liters to find the molarity (M, mol/L).
(0.0981 mol + 0.01024 mol) / 0.5 L = 0.10834 mol / 0.5 L = 0.21668 M
With sig figs it's 0.217 M Cl-.
Question 2:
You need to determine the moles of acetic acid in 20 mL.
Pre-step: Find the MW of acetic acid (CH3COOH). See the problem above for the steps. But the MW of acetic acid is 60 g/mol. You'll also need to convert 250 mL to liters, which is 0.25 L.
Step 1. Convert the mL to g by taking the volume and multiplying by the density.
20 mL acetic acid x 1.049 g/mL acetic acid = 20.98 g acetic acid
Step 2. Convert the g into mol. Do so by taking the mass and dividing by the MW.
20.98 g acetic acid / 60 g/mol acetic acid = 0.350 mol acetic acid
Step 3. Calculate M by taking the mol and diving by final volume of the solution in L.
0.350 mol / 0.25 L = 1.4 M
Question 3:
Step 1. Determine the number of moles in each solution. Do so by multiplying the volume of the solution, in L, by the M.
.03590 L x 1 mol/L = .03590 mol KBr
.060 L x .600 mol/L = .0360 mol KBr
Step 2. Add up the two values to get the total number of moles.
0.03590 + 0.0360 = 0.0719 mol KBr
Step 3. Divide the moles by the volume of the solution in L to get M.
0.0719 mol KBr / 0.050 L = 1.438 M KBr