How would you make up 255 mL of 0.150 M HNO3 from nitric acid that is 68.0% HNO3?
The density of 68.0% HNO3 is 1.41 g/mL.
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someone help me on this plox??
Supposing the "68%" is 68% by weight:
(0.255 L) x (0.150 mol/L) x (63.0130 g/mol) / 0.680 / (1.41 g/ml) = 2.51 ml
The direct way would be to take 2.51 ml of the 68% solution and dilute it to exactly 255 mL.
But as a practical matter, since measuring 2.51 mL accurately might be difficult, it would be useful to make an intermediate dilution of something like 10 to 1, and use the diluted solution to make up the 255 ml of desired concentration.
I agree with the answer by james but I would like to point out two or three things. First, the percent IS by weight. Second, the 68% is never EXACTLY 68%. Third, since you can't easily measure out 2.51 mL AND you can't easily measure out a final solution of 255 mL (and the 68% is just a close 68%), the best you can do is come close to 0.150 M. I realize this is just a practice problem but I wanted to add this information.
To make up 255 mL of 0.150 M HNO3, we need to calculate the volume and mass of the 68.0% HNO3 solution required.
Step 1: Calculate the mass of the 68.0% HNO3 solution required.
To find the mass, we can multiply the volume (in mL) by the density (in g/mL):
Mass = Volume × Density
Mass = 255 mL × 1.41 g/mL
Step 2: Calculate the mass of HNO3 required:
Since the 68.0% HNO3 solution contains 68.0 g of HNO3 in 100 g of the solution, we can calculate the mass of HNO3 required using the equation:
Mass HNO3 = Mass Solution × (68.0 g HNO3 / 100 g Solution)
Mass HNO3 = (255 mL × 1.41 g/mL) × (68.0 g HNO3 / 100 g Solution)
Step 3: Calculate the amount of HNO3 required in moles:
To find the moles of HNO3 required, we need to divide the mass of HNO3 by its molar mass. The molar mass of HNO3 is 63.01 g/mol.
Moles HNO3 = Mass HNO3 / Molar Mass HNO3
Moles HNO3 = [(255 mL × 1.41 g/mL) × (68.0 g HNO3 / 100 g Solution)] / 63.01 g/mol
Step 4: Calculate the volume of 68.0% HNO3 solution required to make 0.150 M HNO3:
Since Molarity (M) = Moles HNO3 / Volume (L), we can rearrange the equation to find the volume:
Volume (L) = Moles HNO3 / Molarity (M)
Volume (L) = [(255 mL × 1.41 g/mL) × (68.0 g HNO3 / 100 g Solution)] / [63.01 g/mol × 0.150 mol/L]
By solving these equations, you can determine the volume of 68.0% HNO3 solution required to make 255 mL of 0.150 M HNO3.