how much of a .2M solution can you make with 340 g of calcium carbonate ( MW=100.0869)
Do I multiply 340g by 1mole/MW
First determine how many mols you have. The molar mass is about 100 so 340g/100 = 3.40 mols.
Then mols = M x L. You know mols and M, solve for L That wll be
3.40 = 2M x L
M = 3.40/2 - 1.20L or 1200 mL.
Yes, you start by 340 x (1/100) = ? which I've done above.
so how would I find the percentage of that solution
To determine the amount of a .2M (molar) solution you can make with 340 g of calcium carbonate (CaCO3), you need to follow a few steps.
First, calculate the number of moles of calcium carbonate using its molar mass.
Molar mass of CaCO3 = 40.08 g/mol (Ca) + 12.01 g/mol (C) + 3(16.00 g/mol) (O) = 100.09 g/mol
Number of moles of CaCO3 = 340 g / 100.09 g/mol = 3.398 moles
Next, using the molarity of the desired solution, you can calculate the volume (in liters) of the solution you can make.
Molarity (M) = moles of solute / volume of solution (in liters)
0.2 M = 3.398 moles / volume of solution (in liters)
Solving for the volume of solution:
Volume of solution (in liters) = 3.398 moles / 0.2 M = 16.99 liters
Therefore, you can make approximately 16.99 liters of a 0.2M calcium carbonate solution using 340 g of calcium carbonate.