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Question

How could someone have calculated the dilution of the HCL from 20.0 cm^3 to 250 cm^3?

Answer 1

Na2CO3 + 2HCl ----------> 2NaCl + CO2 + H2O
1.00 moldm3 of hydrochloric acid.
4.00gdm3 impure sodium carbonate.

20.00cm3 of HCl was added to a 250cm3 volumetric flask, making up the rest of the content with distilled water. which is then added to 25.00cm3 of Na2CO3 through a burette.

Answer 2

The concn of the HCl was 1.00 M. Upon dilution it is 1.00 x (20/250) = ?M

Answer 3

Thank you!!!