Answer :

The mass of HNO₃ present in 6.01 l of aqueous nitric acid which has a ph of 1.37 will be 1.69 grams.

To determine the mass of HNO₃ present in the given solution, we need to use the pH value and the volume of the solution. The concentration of HNO₃ can be calculated from the pH value using the following equation:

[HNO₃] = [tex]10^(-pH)[/tex]

Let's calculate the concentration of HNO₃ first:

[HNO₃] = [tex]10^(-1.37)[/tex] = 0.04466835 M

The molarity of the solution is given as 0.04466835 M, which means there are 0.04466835 moles of HNO₃ per liter of solution.

Next, we need to calculate the number of moles of HNO₃ in the given volume of the solution (6.01 L):

Moles of HNO₃ = Concentration of HNO3 x Volume of Solution

Moles of HNO₃ = 0.04466835 mol/L x 6.01 L

We will calculate the mass of HNO₃ using its molar mass:

Mass of HNO₃ = Moles of HNO₃ x Molar mass of HNO₃

The molar mass of HNO₃ is approximately 63.01 g/mol.

Let's substitute the values and calculate the mass of HNO₃:

Mass of HNO₃ = (0.04466835 mol/L x 6.01 L) x 63.01 g/mol

Mass of HNO₃ = 1.69 grams

Therefore, the mass of HNO₃ present in 6.01 L of the solution is approximately 1.69 grams.

Learn more about nitric acid from the given link:

https://brainly.com/question/29769012

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