Question

230 g of a solution that is 0.65 m in Na2CO3, starting with the solid solute.

mNa2CO3= g,

mH2O = g

Answer 1

Let's assume 1 L of 0.65 m solution of Na2CO3. The weight of this solution = The weight of water + the weight of Na2CO3.

The weight of water will be obviously 1000 gm, as the volume is 1 L. (Here we assume that when you dissolve Na2CO3 in water, the volume does not change. Only the weight changes.)

Now the solution is 0.65 m and volume is 1 L, so the moles of Na2CO3 will be 0.65 moles. The molecular weight of Na2CO3 is 106 gm/mole. Thus the Na2CO3 dissolved will be 106 x 0.65 = 68.9 gm.

So the total weight of the solution will be 1000 + 68.9 = 1068.9 gm.