if i wanted to disolve 106 g of Na2CO3 in enough h20 to make 6.00 L of solution....what is the molarity of this solution?
molarity = moles/L of solution.
moles = g/molar mass = 106/106 = 1 mole
1 mole/6.00 L = ?? M.
To find the molarity of a solution, you need to divide the number of moles of solute by the volume of the solution in liters. In this case, the solute is Na2CO3 (sodium carbonate) and the volume of the solution is 6.00 L.
To determine the number of moles of Na2CO3, you need to convert the given mass of Na2CO3 (106 g) into moles. The molar mass of Na2CO3 can be calculated by adding up the atomic masses of its constituents:
Na: 2 x 22.99 g/mol = 45.98 g/mol
C: 1 x 12.01 g/mol = 12.01 g/mol
O: 3 x 16.00 g/mol = 48.00 g/mol
Total molar mass = 45.98 + 12.01 + 48.00 = 105.99 g/mol (round to 106.00 g/mol)
Now, you can calculate the number of moles:
moles of Na2CO3 = mass / molar mass
moles of Na2CO3 = 106 g / 106.00 g/mol = 1.00 mol
Finally, you can calculate the molarity:
Molarity = moles of solute / volume of solution
Molarity = 1.00 mol / 6.00 L = 0.167 M (rounded to three significant figures)
Therefore, the molarity of the Na2CO3 solution is 0.167 M.