How many grams of solute are present in 705 mL of 0.900 M #KB#r?

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Answer 1 · 2018-05-14T19:15:33

Approx. #75*g#....

By definition....#"concentration"-="moles of solute"/"volume of solution"#

And so to get the #"moles of solute"#, we take the product....

#"moles of solute"="concentration"xx"volume of solution"#..

And so here we gots.......

#n_"KBr"=0.900*mol*L^-1xx705*mLxx10^-3*L*mL^-1#

#=0.6345*mol#...

And thus a MASS of #0.6345*cancel(mol)xx119.0*g*cancel(mol^-1)=??*g#