If you have 435.0 g of water and wish to make a 7.68 M solution of NAOH how many grams of the solute would you have to add to the water that you have?

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Answer 1 · 2018-06-04T21:48:08

Well, I would add the water TO the solute, and then fill up to the mark... That way, we don't fail to make the proper concentration...

It depends on what temperature you are at... if the AVERAGE density of water is #"0.998 g/mL"#, then we have

#435.0 cancel"g" xx "1 mL"/(0.998 cancel"g") = "435.9 mL"#

#=# #"0.4359 L"# #ul"WATER"##", and not solution"# available.

What we want to do is begin with an empty container with the mass of solute already weighed out:

#"0.4359 L solution" xx "7.68 mol"/"L" xx "39.997 g NaOH"/"1 mol"#

#=# #"133.9 g NaOH"#

To properly make this solution, we add water to this magical container that is marked at precisely #"435.9 mL"# and fill it up to that mark.

We will NOT end up using all #"435 g"# of water. Likely, we'll use around #"430 g"# of it.