What mass of solute is present in a #0.250*L# volume of #NaOH# solution, for which #[NaOH]-=2.50*mol*L^-1#?

1 Answer
May 30, 2017

Answer:

Approx. #25*g.................#

Explanation:

First we calculate the molar quantity present in the given volume of #NaOH#........

#"Molarity"="Moles of solute"/"Volume of solution"...........#

And thus #"Moles of solute"="Volume"xx"molarity"#.....

#250*xx10^-3*cancelLxx2.5*mol*cancel(L^-1)=0.625*mol#, with respect to #NaOH#...............

which molar quantity corresponds to a MASS of..........

#0.625*molxx40.0*g*mol^-1=??*g#

You must simply KNOW that #1*mL-=1xx10^-3*L#, i.e. the #"milli"# prefix #-=10^-3#.