A beaker contains 100 grams of 1.20 M NaCl. If you transfer 50% of the solution to another beaker, what is the molarity of the solution remaining in the first beaker?

Answer 1

Think about it a little. Suppose we have a beaker containing a 1 M solution of NaCl. Suppose the beaker contains 1 L of solution; therefore, we know it contains 1 mol/L or 58.5(approximately) g NaCl/1000 mL.
Now let's pour 500 mL out onto the floor. What do we have left in the beaker? We have 0.50 mole NaCl (since we lost half of it) and we have 500 mL left (0.5L since we lost 0.5L) so the concn is not 0.5mol/0.5L and voila, the solution is still 1.0 M. Let's take the same beaker we had at the beginning and pour 900 mL on the floor. That leaves us 100 mL in the beaker(the other 900 is on the floor) and it leaves 0.1 mole NaCl(the other 0.9 is on the floor). So the concn of th solution remaining in the beaker is 0.1mole/0.1L = voila, 1.0M. Think about it again. Do you think simply pouring some of the solution out of one container into another container will change the concn. It's true you lose some of the solute (the NaCl in this case) but it's also true that you lose the same amount of solvent, so the concn remains exactly the same.