nnekadike01 nnekadike01
  • 10-06-2018
  • Chemistry
contestada

How would you prepare 500.0 mL of a 2.00 mol/L solution of HCI from a concentrated solution of 12.0 mol/L?

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Joakim
Joakim Joakim
  • 24-06-2018

c(desired)*V(final) = c(conc)*V(add)

V(add) = c(desired)*V(final) / c(conc) = 2*0.5 / 12 = 0.0833 cm3 = 83.3 mL of 12 M HCl needs to be added.

First add the distilled water: 416,7 mL.

Then add the conc. HCL by using an autopipette 8 times with 10.0 mL, 1 time with 3.0 mL and then 1 time with 300 microliter .

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