Question

Concentrated aqueous nitric acid is 70.4 wt% hno3 and has a concentration of 15.9 m. calculate the volume of concentrated nitric acid that should be diluted to 1.80 l to form a 5.00 m hno3 solution.

Answer 1

This is known that the number of moles of solute will remain the same even on dilution.

The relation between mole and molarity is

`moles=molarityXvolume`

so the moles before and after dilution will be equated as:

`M1V1=M2V2`

Where

M1=initial molarity = 15.9 M

V1= initial volume = ?

M2= Final molarity = 5 M

V2 = final volume = 1.80 L

`Initial volume=(M2V2)/(V1)=(5X1.8)/(M1)=0.566L`

Thus we will take 0.566 L of concentrated nitric acid and will dilute it to 1.8 L.

Answer 2

The volume should be taken from the stock is 0.5335 L = 533.5 ml.

Step-by-step explanation:

• Firstly, we must calculate the molarity of the HNO₃, M = (10pd / Molar mass),
• where p is the weight percent (70.4 %), d is the specific garvity of HNO₃ (d = 1.51 g/cm³) and molar mass = 63.01 g/mole.

• Now, Molarity of the stock M = 10pd/ Molar mass = (10 x 70.4 x 1.51) / (63.01) = 16.87 M.

• To calculate the volume needed to prepare 1.80 L with 5.0 M concentration, we use the rule that the number of millimoles before dilution is equal to the number of millimoles after the dilution.

(MV) before dilution = (MV) after dilution

• The volume should be taken from the stock = (MV) after dilution / M before the dilution = (5.0 M x 1.80 L) / 16.87 M = 0.5335 L = 533.5 ml.