To neutralize 8.3 4 ml of HCl 36.5% by mass and density =

1, 2 g/cm3, the required volume of 5 mol/L NaOH will be?

To determine the required volume of 5 mol/L NaOH to neutralize 8.34 ml of HCl, we need to follow a step-by-step approach. Here's how you can calculate it:

Step 1: Calculate the mass of HCl.
Since we know the density and volume of the HCl solution, we can use the formula:
Mass = Density x Volume

Given:
Density of HCl solution = 1.2 g/cm^3
Volume of HCl solution = 8.34 mL

Mass of HCl = 1.2 g/cm^3 x 8.34 mL

Step 2: Calculate the number of moles of HCl.
To find the number of moles, we need to divide the mass of HCl by its molar mass. The molar mass of HCl is 36.5 g/mol.

Number of moles of HCl = Mass of HCl / Molar mass of HCl
Number of moles of HCl = (1.2 g/cm^3 x 8.34 mL) / 36.5 g/mol

Step 3: Calculate the required volume of NaOH.
The reaction between HCl and NaOH is 1:1. This means that one mole of HCl reacts with one mole of NaOH. Therefore, the moles of NaOH needed to neutralize the moles of HCl will be the same.

Number of moles of NaOH = Number of moles of HCl

Step 4: Calculate the required volume of NaOH.
To find the required volume of NaOH, we divide the number of moles of NaOH by its concentration (mol/L). Given that the concentration of NaOH is 5 mol/L, we can use the formula:

Volume of NaOH = Number of moles of NaOH / Concentration of NaOH

Volume of NaOH = Number of moles of HCl / 5 mol/L

Now that we have the necessary information and formulas, we can substitute the values and calculate:

Volume of NaOH = [(1.2 g/cm^3 x 8.34 mL) / 36.5 g/mol] / 5 mol/L

By evaluating this expression, you will get the required volume of NaOH in liters.