A #295mL# volume of solution that is #1.00molL^-1# with respect to #"ammonium chloride"# is diluted with a #2.25L# volume of fresh solvent. What is the resultant #[NH_4Cl]#?

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Answer 1 · 2016-08-14T17:18:01

Approx. #0.116*mol*L^-1#

As with all these sorts of problems,

#"concentration"="moles of solute"/"volume of solution"#

So we just work out moles and volume appropriately:

#=# #(295*mLxx10^-3*L*mL^-1xx1.00*mol*L^-1)/(0.295*L+2.25*L)#

#=# #??mol*L^-1# with respect to #NH_4Cl#.

The concentration makes sense, as you dilute the initial concentration almost 10fold.

A #295*mL# volume of solution that is #1.00*mol*L^-1# with respect to #"ammonium chloride"# is diluted with a #2.25*L# volume of fresh solvent. What is the resultant #[NH_4Cl]#?