Less quantitative dilutions can also be done without a volumetric flask, The pre

Question

Less quantitative dilutions can also be done without a volumetric flask, The precision of concentration is dependent on how precise the volumes are transferred. For example a dilute solution of HCl can be made using a precise amount of 5.00 M HCl and distilled water. In this case the total volume of dilute HCl (V2) is the total volume of 5.00 M HCl plus the volume of water. Determine the volume of water and the volume (V1) of 5.00 M (M1) HCl in mL needed to prepare 5.00 mL (V2) of 3.40 M (M2) HCl solution: Volume of 5.00 M HCl ==>
Volume of distilled water delivered using a graduated cylinder, Mohr pipet, or pipettor ==>

Explanation / Answer

For this, we apply dilution equation:

M1V1 = M2V2

5M (V1) = 3.4M (.005L)

V1 = .0034L = 3.4 mL of 5M HCl

Moles of HCl = .0034 L * 5 mol/L = .017 mol

HCl molar mass = 36.46 g/mol * .017 mol = .61982 g

HCl density = .61982 g / 1.1g/ml = .5634 ml

Water volume = 3.4 - .5634 = 2.8365 ml

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