How is a #35%# #"w/w"# solution of a #200*g# mass of sodium chloride in water prepared?

1 Answer
anor277
Jul 25, 2017

About #70*g# of salt, and #130*mL# water.........

Explanation:

#"Concentration"="Mass of solute"/"Mass of solute + mass of water"xx100%#

And thus #35%="Mass of solute"/(200*g)#....

And so #"mass of solute"-=35%xx200*g=70*g#.

And the volume of the solution would change only very marginally, even upon the addition of such a mass of solute.

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