Question

1. you can purchase nitric acid in a concentrated form that is 70.3% hno₃ by mass and has a density of 1.41 g ml⁻¹. describe exactly how you would prepare 1.15 l of 0.100 mol l⁻¹ hno₃ from the concentrated solution.

Explanation:

Step1: Calculate the molarity of concentrated \(HNO_3\)

Let's assume we have \(1000\space mL\) of the concentrated solution.
The mass of the solution \(m=
ho V\), where \(
ho = 1.41\space g/mL\) and \(V = 1000\space mL\). So \(m=1.41\times1000 = 1410\space g\)
The mass of \(HNO_3\) in the solution is \(m_{HNO_3}=0.703\times1410\space g\)
The molar mass of \(HNO_3\) is \(M = 63.01\space g/mol\)
The number of moles of \(HNO_3\), \(n=\frac{0.703\times1410}{63.01}\space mol\)
The molarity \(C_1=\frac{n}{V}\) (where \(V = 1\space L\))

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Step2: Use the dilution formula \(C_1V_1 = C_2V_2\)

We know \(C_2=0.100\space mol/L\), \(V_2 = 1.15\space L\) and \(C_1\approx15.73\space mol/L\)

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Step3: Describe the dilution process

Measure out \(7.31\space mL\) of the concentrated \(HNO_3\) (using a graduated cylinder). Transfer it to a \(1.15\space L\) volumetric flask. Add water to the mark on the volumetric flask while swirling the flask to ensure uniform mixing.

Answer:

Measure out \(7.31\space mL\) of the concentrated \(HNO_3\) (using a graduated cylinder). Transfer it to a \(1.15\space L\) volumetric flask. Add water to the mark on the volumetric flask while swirling the flask to ensure uniform mixing.