A flask containing 5ml of 3M hcl solution required 14.45ml of 1M NAOH for titration. How many moles of hcl are present in the solution. The density of 3M hcl is 1.05g/ml

My calculation
0.005 x 3 = 0.015m
0.01445 x 1 = 0.01445m

Who helps me to solve it

You have two answers. Which is the right one?

Its 0.01445 x 1 = 0.01445 mols HCl because it took 0.01445 mols NaOH to neutralize it. You obviously don't have 5 mL of 3M HCl.

To solve this problem, you need to use the concept of stoichiometry and the molarity of the given solutions. Here's how to do it step by step:

1. Start by converting the volume of the 3M HCl solution from milliliters (ml) to grams (g) using its density. The density of the solution is given as 1.05 g/ml. Multiply the volume (5 ml) by the density to get the mass of the solution.

Mass of HCl solution = Volume of solution x Density
Mass of HCl solution = 5 ml x 1.05 g/ml = 5.25 g

2. Convert the mass of the HCl solution to moles using its molarity. The molarity of the HCl solution is given as 3M. To convert grams to moles, divide the mass by the molar mass of HCl.

Molar mass of HCl = 1 g/mol + 35.45 g/mol = 36.45 g/mol

Moles of HCl solution = Mass of HCl solution / Molar mass of HCl
Moles of HCl solution = 5.25 g / 36.45 g/mol = 0.1436 mol

3. Use the stoichiometric ratio between HCl and NaOH to determine the number of moles of HCl present in the solution. From the balanced chemical equation:

HCl + NaOH → NaCl + H2O

The stoichiometric ratio between HCl and NaOH is 1:1. This means that for every 1 mole of HCl, 1 mole of NaOH is required for complete reaction.

Since the given 1M NaOH solution is used in the titration, the moles of NaOH used (0.01445 mol) also represents the moles of HCl present in the solution.

So, the moles of HCl present in the solution = 0.01445 mol

Therefore, the number of moles of HCl present in the solution is 0.01445 mol.