What is the molarity of the resulting saline solution when you mix 1 mole HCl with 1 mole NaOH? I'll provide some basic information that I know, but could somebody tell me how to get the answer using the data?

The balanced reaction is as follows: HCl+NaOH->H2O+NaCl

Molar Mass is as follows

1 mole HCl=36.4606 grams
1 mole NaOH=39.9971 grams
1 mole H2O=18.01524 grams
1 mole NaCl=58.442 grams

1 gram H2O=1 ml H2O

The molarity of the solution equals the number of moles of solute (NaCl)in the solution divided by the number of liters of the solution.

Could somebody tell me how to use the information?

You need a final volume before you can answer this question. We can make some assumptions and get at the answer but technically we need the final volume.

1 mol HCl + 1 mol NaOH will produce 1 mol NaCl. Since the only volume comes from the HCl solution (assuming the solid NaOH doesn't change the volume) we can make some guesses and be close, probably, but we must guess at the volume if you don't have it in the problem.

To find the molarity of the resulting saline solution, we need to determine the number of moles of NaCl produced by the reaction.

First, calculate the number of moles of NaCl produced by the reaction. According to the balanced equation, the stoichiometry is 1:1 between HCl and NaCl. This means that for every mole of HCl, we get one mole of NaCl.

Since we have mixed 1 mole of HCl with 1 mole of NaOH, we know that there will be 1 mole of NaCl produced.

Next, we calculate the mass of NaCl produced using its molar mass (58.442 grams/mole):

Mass of NaCl = 1 mole NaCl * 58.442 grams/mole = 58.442 grams

Since we have 58.442 grams of NaCl, and we know that 1 gram of water is equal to 1 mL of water, we can now determine the volume of the solution in liters.

Volume of solution = Mass of NaCl (in grams) = 58.442 grams = 58.442 mL
Density of water 1 g/mL

Finally, we can calculate the molarity of the solution by dividing the number of moles of NaCl by the volume of the solution in liters.

Molarity = Moles of NaCl / Volume of solution (in liters)
= 1 mole NaCl / 58.442 mL (converted to liters)
= 1 mole NaCl / 0.058442 L
= 17.1171 moles/Liter

Therefore, the molarity of the resulting saline solution is approximately 17.1 M.