If 25.0 g of #K_2CO_3#, potassium carbonate, are dissolved in #450 cm^3# of solution, what is the molarity?

Question

Cameron Andrews · Accepted Answer

Approx. #0.4*mol*L^-1# #"Molarity"# #=# #"Moles of potassium carbonate (mol)"/"Volume of solution (L)"# #=((25*g)/(138.21*g*mol^-1))/(0.450*L)~=0.4*mol*L^-1# Note that when we do the calculation this wa we automatically get appropriate units. We seek the molarity, which should properly have units of #mol*L^-1#, and lo and behold, the expression gives us such units as the grams cancel out, and the #mol^-1# is inverted to give #1/(1/(mol))=mol#.......

Axel Alvarado · Answer

undefined To find the molarity (M) of the solution, first calculate the number of moles of K₂CO₃ using its molar mass (M₁):
M₁(K₂CO₃) = 2 * M(K) + M(C) + 3 * M(O) = 2 * 39.10 g/mol + 12.01 g/mol + 3 * 16.00 g/mol = 138.21 g/mol

Number of moles = mass / molar mass = 25.0 g / 138.21 g/mol ≈ 0.181 moles

Then, use the formula for molarity:
Molarity (M) = moles / volume (in liters)
Volume = 450 cm³ = 450 cm³ * (1 L / 1000 cm³) = 0.450 L

Molarity (M) = 0.181 moles / 0.450 L ≈ 0.402 M

The molarity of the solution is approximately 0.402 M.