in a .05M (point zero five molar) solution, you are saying you add .05 moles of your solute (the Na2Co3 in this instance) to one liter of solvent (in this case, water).
What part of a liter is 250 ml? (Hint: 1000 ml = 1 liter.) Once you have that proportion, then you can find the proportion of moles of sodium carbonate to use for 250 ml of water.
250 ml * X = 1000 ml
(X is your proportion)
so now, if you add .05 moles of solute to one liter of water, you multiply your moles by the proportion you got in the above equation. That is, X. This will equal .05 moles that you have in a liter of solution.
(number of moles of Na2CO3) * X = .05 moles Na2CO3 in a liter of water
So when you reduce the amount of solvent, you reduce the amount of solute by the same amount in order to keep the same molarity.
Now, here's the hard part: number of moles implies how much mass?
First, go to the periodic table and get the atomic masses of the elements in sodium carbonate. Multiply the atomic masses of each by the number of atoms of that element that appear in the molecule:
2 atoms Na * atomic mass of Na + 3 atoms of O * atomic mass of O + 1 atom of C * atomic mass of C = your total atomic mass of sodium carbonate
When you put g for grams after the atomic mass of sodium carbonate, you are expressing the amount of mass of one mole of sodium carbonate.
For an example, take molecular oxygen: oxygen as a pure gas is present as two atoms per molecule. So a molecular mass of one oxygen molecule is always twice the mass of one atom of oxygen. Oxygen has an atomic mass of 16. Two atoms of oxygen has a molecular mass of 32. Thus, a mole of oxygen has a mass of 32g.
So a mole of one compound is the molecular mass of that compound with g after it.
A mole of carbon (C) is 12g
A mole of hydrogen (H2) is 1g
A mole of iron (Fe) is 26g
A mole of Uranium 235 (U235) is 235g
A mole of potassium (K) is 19g
Look at the periodic table and see how these relate.
Once you know the mass of one mole of your solute, then you can multiply that by how many moles you need for your solution.