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Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: thirdbasekid24 on December 07, 2004, 01:33:56 AM

Title: Dilutions
Post by: thirdbasekid24 on December 07, 2004, 01:33:56 AM

Determine the number of mL of .494 M ammonium sulfite solution that are required to prepare 85.4 mL of a solution that is .179 M in ammonium ion?

i am confused on the set up... i went .0854 L x .179 M which = .0152 and divided by .494... and i got an answer that was on the worksheet but it wasnt correct...can somebody tell me how to set it up correctly it would be muchly appriciated

Title: Re:Dilutions
Post by: AWK on December 07, 2004, 01:45:31 AM

Amonium sulfite (NH4)2S shows doubled concentration of ammonium ions in relation to its molar concetration.