Question

What is the molarity of a solution prepared by dissolving 543g ofCa(NO3)2 into 355ml of water?

Answer 1

What is the final concentration (M) of a solution prepared by diluting 50.0ml of a6.00m KCl solution to a volume of 2.00L?

Answer 2

First, you need to calculate the number of moles of Ca(NO3)2 in 543g:

molar mass of Ca(NO3)2 = 164 g/mol
moles of Ca(NO3)2 = mass / molar mass
moles of Ca(NO3)2 = 543g / 164 g/mol = 3.31 mol

Next, you need to calculate the volume of the solution in liters:

355 ml = 0.355 L

Finally, you can calculate the molarity of the solution:

molarity = moles of solute / volume of solution in liters
molarity = 3.31 mol / 0.355 L = 9.31 M

Therefore, the molarity of the solution is 9.31 M.

Answer 3

To find the final concentration of the solution, you can use the equation:

C1V1 = C2V2

where C1 is the initial concentration, V1 is the initial volume, C2 is the final concentration, and V2 is the final volume.

Substituting the given values:

C1 = 6.00 M
V1 = 50.0 mL = 0.0500 L
V2 = 2.00 L

Solving for C2:

C2 = (C1V1) / V2
C2 = (6.00 M x 0.0500 L) / 2.00 L
C2 = 0.150 M

Therefore, the final concentration of the solution is 0.150 M.