To neutralize 48 ml of Na2CO3 solution, 30 ml of N / 10 HCl is required. We need to find out how much water should be mixed with 50 ml of this solution to obtain a solution with a concentration of N / 40.

Let's break down the problem into steps:

Step 1: Calculate the number of moles of Na2CO3 in the given solution.
- The concentration of the HCl solution is N / 10, which means it contains 1/10th of a mole of HCl per liter.
- As we have 30 ml of the solution, the number of moles of HCl can be calculated as follows:
Moles of HCl = (30 ml / 1000) * (N / 10) = 0.003 N moles

- Since HCl and Na2CO3 react in a 1:1 ratio, the number of moles of Na2CO3 in the solution is also 0.003 N moles.

Step 2: Calculate the concentration of the Na2CO3 solution.
- The volume of the Na2CO3 solution is given as 48 ml.
- To convert this to liters, we divide by 1000:
Volume of Na2CO3 solution = 48 ml / 1000 = 0.048 L

- The concentration of the Na2CO3 solution can be calculated as follows:
Concentration = Number of moles / Volume
Concentration = (0.003 N moles) / (0.048 L) = 0.0625 N

Step 3: Calculate the volume of water required to dilute the solution.
- Let's assume the final volume of the diluted solution is V ml.
- The concentration of the diluted solution is N / 40, which means it contains 1/40th of a mole of Na2CO3 per liter.
- The number of moles of Na2CO3 in the diluted solution can be calculated as follows:
Moles of Na2CO3 = (V ml / 1000) * (N / 40) = V * N / 40000 moles

- Since the number of moles of Na2CO3 remains the same before and after dilution, we can equate the number of moles in the original solution to the number of moles in the diluted solution:
0.003 N moles = V * N / 40000 moles

- Solving for V, we find:
V = (0.003 N moles * 40000) / N = 120 ml

Therefore, to obtain a solution with a concentration of N / 40, 120 ml of water should be mixed with the 50 ml of the given Na2CO3 solution.

75 ml