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How could someone have calculated the dilution of the HCL from 20.0 cm^3 to 250 cm^3?
How could someone have calculated the dilution of the HCL from 20.0 cm^3 to 250 cm^3?
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Answered by
?
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.
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.
Answered by
DrBob222
The concn of the HCl was 1.00 M. Upon dilution it is 1.00 x (20/250) = ?M
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Thank you!!!
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