in what ratio 0.2M NaCl and 0.1M CaCl2 solutions are to be mixed so that in the resulting solution, the concentration of negative ions is 50% greater than the concentration of positive ions?

Answer 1

I think what you want to do is this.
Let's take a volume (choose anything but I'll pick 100 mL of the 0.2M NaCl) and calculate the amount of CaCl2 needed to meet the problem.
100 mL of 0.2M NaCl + xmL of 0.1M CaCl2, we have
millimols + charge = 100*0.2 = 20 for Na in NaCl and 0.1x for Ca in CaCl2 for total of 20+0.1x and you want to multiply this by 1.5 so it will be 50% higher than the negative charge.

mmols negative charge is 100*0.2 = 20 for Cl in NaCl and 0.2x for the Cl in CaCl2 for total of 20+0.2x. Set these equal for the following:
1.5(20+0.1x) = 20+0.2x and solve for x = mL of 0.1M CaCl2 solution. You can go through this and you should, if for no other reason than to confirm my numbers. I obtained 200 mL of the CaCl2 so the ratio is 100 mL NaCl/200 mL CaCl2 or a ratio of 1 NaCl solution/2 CaCl2 solution. If I were you I would then plug these volumes back into the original solution and confirm that the negative charge = 50% greater than the positive charge. You may even want to try difference numbers, such as 500 mL and 1000 mL (still in the ratio of 1:2) and see if the equation holds.